🧠:<think>Okay, so I have this problem here where I need to figure out which function among options A, B, C, and D is both even and increasing in the interval (1,2). Hmm, let me break this down step by step.First, I remember that an even function satisfies the condition f(-x) = f(x) for all x in its domain. So, I need to check each function to see if it's even. Then, I also need to determine if the function is increasing in the interval (1,2). That means, for any two points x1 and x2 in (1,2) where x1 < x2, the function should satisfy f(x1) < f(x2). Let me go through each option one by one.Option A: y = cos(2x)Alright, starting with option A. I know that cosine is an even function because cos(-x) = cos(x). But here, it's cos(2x). Let me check if this is still even. Let's substitute -x into the function:f(-x) = cos(2*(-x)) = cos(-2x) = cos(2x) because cosine is even. So, yes, cos(2x) is also even. So, A is even. Good.Now, is it increasing in (1,2)? Hmm, to check if a function is increasing, I can look at its derivative. The derivative of cos(2x) is -2 sin(2x). So, f'(x) = -2 sin(2x). I need to see if this derivative is positive in (1,2). Let's think about the values of sin(2x) in that interval. When x is between 1 and 2, 2x is between 2 and 4 radians. I remember that sin(π) is 0, and π is approximately 3.14. So, 2 radians is a bit less than π, and 4 radians is a bit more than π. So, in the interval (2,4), sin(2x) will be positive from 2 to π (which is about 3.14) and negative from π to 4. Wait, so from x=1 to x=2, 2x goes from 2 to 4. So, sin(2x) is positive from x=1 to x=π/2 (which is about 1.57) and negative from x=π/2 to x=2. But wait, hold on, 2x from 2 to 4, so sin(2x) is positive from 2 to π (which is about 3.14) and negative from π to 4. So, in terms of x, 2x=2 corresponds to x=1, 2x=π corresponds to x=π/2≈1.57, and 2x=4 corresponds to x=2. So, sin(2x) is positive from x=1 to x≈1.57 and negative from x≈1.57 to x=2. Therefore, the derivative f'(x) = -2 sin(2x) will be negative when sin(2x) is positive and positive when sin(2x) is negative. So, from x=1 to x≈1.57, f'(x) is negative, meaning the function is decreasing, and from x≈1.57 to x=2, f'(x) is positive, meaning the function is increasing. But the question is asking if the function is increasing in the entire interval (1,2). Since it's decreasing in part of that interval and increasing in another part, it's not strictly increasing throughout (1,2). So, option A is even but not increasing in the entire interval (1,2). So, A is out.Option B: y = log₂|x|Moving on to option B: y = log base 2 of |x|. Let me check if this is even. f(-x) = log₂| -x | = log₂|x| = f(x). So, yes, it's even. Good.Now, is it increasing in (1,2)? Well, for x > 0, log₂|x| is just log₂x. The logarithm function with base greater than 1 is increasing. So, log₂x is increasing for x > 0. Therefore, in the interval (1,2), it's increasing. So, B is both even and increasing in (1,2). Hmm, that seems to fit the criteria. But let me check the other options just to be thorough.Option C: y = (e^x - e^{-x}) / 2Option C is y = (e^x - e^{-x}) / 2. I recognize this as the definition of sinh(x), the hyperbolic sine function. Is sinh(x) even? Let me check f(-x):f(-x) = (e^{-x} - e^{x}) / 2 = - (e^x - e^{-x}) / 2 = -f(x). So, it's an odd function, not even. Therefore, C is not even. So, we can eliminate C.Option D: y = x³ + 1Option D is y = x³ + 1. Let's check if it's even.f(-x) = (-x)^3 + 1 = -x³ + 1. Compare this to f(x) = x³ + 1. So, f(-x) = -x³ + 1, which is not equal to f(x) unless x³ = 0, which isn't generally true. So, D is not even. Therefore, D is out.So, after going through all options, only option B is both even and increasing in the interval (1,2). But just to make sure I didn't make a mistake, let me recap:- A is even but not increasing throughout (1,2).- B is even and increasing in (1,2).- C is odd, not even.- D is not even.Yep, that seems correct. So, the answer should be B.Final Answerboxed{B}</think>

🧠:<think>Alright, so I've got this equation here: (5 + x)/(7 + x) equals (2 + x²)/(4 + x). I need to find the value of x that makes these two fractions equal. Hmm, okay, let's see how to approach this.First off, when I see two fractions set equal to each other, my initial thought is to cross-multiply to eliminate the denominators. That usually makes things simpler. So, if I cross-multiply, I should get (5 + x)(4 + x) equals (7 + x)(2 + x²). Let me write that down:(5 + x)(4 + x) = (7 + x)(2 + x²)Alright, now I need to expand both sides to simplify the equation. Let's start with the left side. Multiplying (5 + x) by (4 + x) should give me:5*4 + 5*x + x*4 + x*x = 20 + 5x + 4x + x²Combining like terms, that becomes:20 + 9x + x²Okay, so the left side simplifies to x² + 9x + 20.Now, let's tackle the right side: (7 + x)(2 + x²). I'll distribute each term in the first parenthesis across the second one.7*2 + 7*x² + x*2 + x*x² = 14 + 7x² + 2x + x³So, the right side simplifies to x³ + 7x² + 2x + 14.Now, putting it all together, the equation becomes:x² + 9x + 20 = x³ + 7x² + 2x + 14Hmm, now I need to get all terms on one side to set the equation to zero. Let's subtract the left side from both sides:0 = x³ + 7x² + 2x + 14 - x² - 9x - 20Simplifying that, combine like terms:x³ + (7x² - x²) + (2x - 9x) + (14 - 20) = 0Which simplifies to:x³ + 6x² - 7x - 6 = 0Wait, that doesn't look quite right. Let me double-check my subtraction:From the right side: x³ + 7x² + 2x + 14Subtracting the left side: -x² -9x -20So, x³ + (7x² - x²) = x³ + 6x²Then, 2x - 9x = -7xAnd 14 - 20 = -6So, yes, the equation is:x³ + 6x² - 7x - 6 = 0Hmm, okay, so now I have a cubic equation. Solving cubic equations can be tricky, but maybe I can factor this. Let's see if there are any rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term divided by factors of the leading coefficient. Here, the constant term is -6, and the leading coefficient is 1, so possible roots are ±1, ±2, ±3, ±6.Let me test x = 1:1³ + 6(1)² - 7(1) - 6 = 1 + 6 - 7 - 6 = -6 ≠ 0Not a root.x = -1:(-1)³ + 6(-1)² - 7(-1) - 6 = -1 + 6 + 7 - 6 = 6 ≠ 0Not a root.x = 2:2³ + 6(2)² - 7(2) - 6 = 8 + 24 - 14 - 6 = 12 ≠ 0Not a root.x = -2:(-2)³ + 6(-2)² - 7(-2) - 6 = -8 + 24 + 14 - 6 = 24 ≠ 0Wait, that's 24, not zero. Hmm, maybe I miscalculated.Wait, (-2)^3 is -8, 6*(-2)^2 is 6*4=24, -7*(-2)=14, and -6. So, -8 +24 is 16, plus 14 is 30, minus 6 is 24. Yeah, that's correct. Not zero.x = 3:3³ + 6(3)² - 7(3) - 6 = 27 + 54 - 21 - 6 = 54 ≠ 0Nope.x = -3:(-3)^3 + 6*(-3)^2 - 7*(-3) - 6 = -27 + 54 + 21 - 6 = 42 ≠ 0Hmm, not zero either.x = 6:6³ + 6(6)² - 7(6) - 6 = 216 + 216 - 42 - 6 = 384 ≠ 0Way too big.x = -6:(-6)^3 + 6*(-6)^2 - 7*(-6) - 6 = -216 + 216 + 42 - 6 = 36 ≠ 0Still not zero.Wait, did I make a mistake earlier? Let me check my equation again.Original equation after cross-multiplying:(5 + x)(4 + x) = (7 + x)(2 + x²)Left side: 5*4 + 5x + 4x + x² = 20 + 9x + x²Right side: 7*2 + 7x² + 2x + x³ = 14 + 7x² + 2x + x³So, moving everything to one side:x³ + 7x² + 2x + 14 - x² - 9x - 20 = 0Which is:x³ + 6x² -7x -6 = 0Wait, earlier I thought it was x³ +6x² -9x -6=0, but actually, it's -7x. So, that changes things.So, the equation is x³ +6x² -7x -6=0.Let me try the possible roots again.x=1:1 +6 -7 -6= -6≠0x=-1:-1 +6 +7 -6=6≠0x=2:8 +24 -14 -6=12≠0x=-2:-8 +24 +14 -6=24≠0x=3:27 +54 -21 -6=54≠0x=-3:-27 +54 +21 -6=42≠0x=6:216 +216 -42 -6=384≠0x=-6:-216 +216 +42 -6=36≠0Hmm, none of the possible rational roots are working. Did I make a mistake in expanding?Wait, let me double-check the expansion.Left side: (5 + x)(4 + x) = 5*4 +5x +x*4 +x*x=20 +5x +4x +x²=20 +9x +x². That seems correct.Right side: (7 + x)(2 + x²)=7*2 +7x² +x*2 +x*x²=14 +7x² +2x +x³. That also seems correct.So, moving everything to one side:x³ +7x² +2x +14 -x² -9x -20= x³ +6x² -7x -6=0Yes, that's correct.Hmm, since none of the rational roots are working, maybe I need to factor this differently or use another method.Alternatively, perhaps I made a mistake in the initial cross-multiplication. Let me check that again.Original equation:(5 + x)/(7 + x) = (2 + x²)/(4 + x)Cross-multiplying: (5 + x)(4 + x) = (7 + x)(2 + x²)Yes, that's correct.Wait, maybe I can factor the cubic equation x³ +6x² -7x -6=0.Let me try grouping:x³ +6x² -7x -6 = (x³ +6x²) + (-7x -6) = x²(x +6) - (7x +6)Hmm, that doesn't seem to help much.Alternatively, maybe factor by grouping differently:x³ -7x +6x² -6 = x(x² -7) +6(x² -1)Hmm, x² -7 doesn't factor nicely, and x² -1 is (x-1)(x+1). Not sure if that helps.Alternatively, maybe use synthetic division to find a root.Since none of the rational roots worked, perhaps I need to consider that this cubic might have irrational or complex roots, which complicates things.Wait, maybe I made a mistake in the cross-multiplication step. Let me double-check.(5 + x)/(7 + x) = (2 + x²)/(4 + x)Cross-multiplying: (5 + x)(4 + x) = (7 + x)(2 + x²)Yes, that's correct.Wait, perhaps I can try to factor the cubic equation differently. Let me see.x³ +6x² -7x -6=0Let me try to factor out an x:x(x² +6x -7) -6=0Hmm, not helpful.Alternatively, maybe factor as (x + a)(x² + bx + c)=0Let me try to find a, b, c such that:(x + a)(x² + bx + c) = x³ + (a + b)x² + (ab + c)x + acComparing to x³ +6x² -7x -6, we have:a + b =6ab + c =-7ac =-6So, from ac=-6, possible integer pairs for (a,c) are (1,-6), (-1,6), (2,-3), (-2,3), (3,-2), (-3,2), (6,-1), (-6,1)Let's try a=3, c=-2:Then, a + b=6 => 3 + b=6 => b=3ab + c=3*3 + (-2)=9 -2=7≠-7Not matching.Next, a=2, c=-3:a + b=6 =>2 + b=6 =>b=4ab + c=2*4 + (-3)=8 -3=5≠-7Nope.a=1, c=-6:a + b=6 =>1 + b=6 =>b=5ab + c=1*5 + (-6)=5 -6=-1≠-7No.a=-1, c=6:a + b=6 =>-1 + b=6 =>b=7ab + c=(-1)*7 +6=-7 +6=-1≠-7No.a=-2, c=3:a + b=6 =>-2 + b=6 =>b=8ab + c=(-2)*8 +3=-16 +3=-13≠-7No.a=-3, c=2:a + b=6 =>-3 + b=6 =>b=9ab + c=(-3)*9 +2=-27 +2=-25≠-7No.a=6, c=-1:a + b=6 =>6 + b=6 =>b=0ab + c=6*0 + (-1)=0 -1=-1≠-7No.a=-6, c=1:a + b=6 =>-6 + b=6 =>b=12ab + c=(-6)*12 +1=-72 +1=-71≠-7No.Hmm, none of these are working. Maybe the cubic doesn't factor nicely, and I need to use the cubic formula or numerical methods. But that seems complicated for a beginner.Wait, maybe I made a mistake earlier. Let me check the original equation again.(5 + x)/(7 + x) = (2 + x²)/(4 + x)Cross-multiplying: (5 + x)(4 + x) = (7 + x)(2 + x²)Yes, that's correct.Wait, perhaps I can simplify the equation differently. Let me try to bring everything to one side:(5 + x)/(7 + x) - (2 + x²)/(4 + x) = 0To combine these fractions, I need a common denominator, which would be (7 + x)(4 + x). So:[(5 + x)(4 + x) - (2 + x²)(7 + x)] / [(7 + x)(4 + x)] = 0Which means the numerator must be zero:(5 + x)(4 + x) - (2 + x²)(7 + x) = 0Wait, that's the same as before, leading to the same cubic equation. So, I'm back to where I started.Hmm, maybe I need to consider that the cubic equation has one real root and two complex roots, but since we're looking for real solutions, perhaps there's only one real root. But earlier attempts didn't find any rational roots, so maybe I need to use the rational root theorem more carefully or consider that I might have made a mistake in the expansion.Wait, let me try expanding again to make sure.Left side: (5 + x)(4 + x) = 5*4 +5x +x*4 +x*x=20 +5x +4x +x²=20 +9x +x²Right side: (7 + x)(2 + x²)=7*2 +7x² +x*2 +x*x²=14 +7x² +2x +x³So, moving everything to one side:x³ +7x² +2x +14 -x² -9x -20= x³ +6x² -7x -6=0Yes, that's correct.Wait, maybe I can try factoring by grouping again, but differently.x³ +6x² -7x -6= x²(x +6) - (7x +6)Hmm, not helpful.Alternatively, maybe factor as (x + a)(x² + bx + c)=0, but as I tried earlier, it didn't work.Alternatively, maybe use the cubic formula, but that's quite involved.Alternatively, perhaps I can graph the function f(x)=x³ +6x² -7x -6 and see where it crosses zero.Alternatively, maybe use the Newton-Raphson method to approximate the root.But since this is a problem likely intended to have integer solutions, perhaps I made a mistake in the earlier steps.Wait, let me check the original equation again:(5 + x)/(7 + x) = (2 + x²)/(4 + x)Cross-multiplying: (5 + x)(4 + x) = (7 + x)(2 + x²)Yes, that's correct.Wait, perhaps I can try plugging in x=1 again:Left side: (5+1)/(7+1)=6/8=3/4Right side: (2+1)/(4+1)=3/53/4 ≠ 3/5, so x=1 is not a solution.x=-2:Left side: (5-2)/(7-2)=3/5Right side: (2 +4)/(4-2)=6/2=33/5 ≠3x=-3:Left side: (5-3)/(7-3)=2/4=1/2Right side: (2 +9)/(4-3)=11/1=111/2 ≠11Hmm, none of these are working. Wait, but in the initial problem, the assistant's answer was x=1, -2, -3, but when I plug them in, they don't satisfy the equation. So, perhaps there was a mistake in the assistant's solution.Wait, let me check the assistant's solution:They started with (5 + x)/(7 + x) = (2 + x²)/(4 + x)Cross-multiplied to get (5 + x)(4 + x) = (7 + x)(2 + x²)Expanded to 20 +5x +4x +x²=14 +7x² +2x +x³Simplified to x³ +6x² -9x -6=0Wait, in my calculation, I got x³ +6x² -7x -6=0, but the assistant got x³ +6x² -9x -6=0. So, there's a discrepancy.Let me check the assistant's expansion:(5 + x)(4 + x)=20 +5x +4x +x²=20 +9x +x²(7 + x)(2 +x²)=14 +7x² +2x +x³So, moving everything to one side:x³ +7x² +2x +14 -x² -9x -20= x³ +6x² -7x -6=0But the assistant wrote:x³ +6x² -9x -6=0So, the assistant must have made a mistake in the expansion, perhaps in the coefficients.Wait, in the assistant's solution, they wrote:20 +5x +4x +x²=14 +7x² +2x +x³Then, simplifying and collecting like terms:x³ +6x² -9x -6=0But in reality, it should be:Left side: x² +9x +20Right side: x³ +7x² +2x +14So, moving everything to one side:x³ +7x² +2x +14 -x² -9x -20= x³ +6x² -7x -6=0So, the assistant incorrectly wrote -9x instead of -7x.Therefore, the correct equation is x³ +6x² -7x -6=0, not x³ +6x² -9x -6=0.So, the assistant's factoring was based on an incorrect equation, leading to incorrect roots.Therefore, the correct equation is x³ +6x² -7x -6=0, which doesn't factor nicely with rational roots, as I saw earlier.So, perhaps the original problem was intended to have a different setup, or perhaps there's a typo.Alternatively, maybe I can try to factor the correct cubic equation.Let me try to factor x³ +6x² -7x -6.Let me try grouping:x³ +6x² -7x -6 = x²(x +6) - (7x +6)Hmm, not helpful.Alternatively, maybe factor as (x + a)(x² + bx + c)=0We have:(x + a)(x² + bx + c) = x³ + (a + b)x² + (ab + c)x + acComparing to x³ +6x² -7x -6, we have:a + b =6ab + c =-7ac =-6So, from ac=-6, possible integer pairs for (a,c) are (1,-6), (-1,6), (2,-3), (-2,3), (3,-2), (-3,2), (6,-1), (-6,1)Let's try a=3, c=-2:Then, a + b=6 =>3 + b=6 =>b=3ab + c=3*3 + (-2)=9 -2=7≠-7No.a=2, c=-3:a + b=6 =>2 + b=6 =>b=4ab + c=2*4 + (-3)=8 -3=5≠-7No.a=1, c=-6:a + b=6 =>1 + b=6 =>b=5ab + c=1*5 + (-6)=5 -6=-1≠-7No.a=-1, c=6:a + b=6 =>-1 + b=6 =>b=7ab + c=(-1)*7 +6=-7 +6=-1≠-7No.a=-2, c=3:a + b=6 =>-2 + b=6 =>b=8ab + c=(-2)*8 +3=-16 +3=-13≠-7No.a=-3, c=2:a + b=6 =>-3 + b=6 =>b=9ab + c=(-3)*9 +2=-27 +2=-25≠-7No.a=6, c=-1:a + b=6 =>6 + b=6 =>b=0ab + c=6*0 + (-1)=0 -1=-1≠-7No.a=-6, c=1:a + b=6 =>-6 + b=6 =>b=12ab + c=(-6)*12 +1=-72 +1=-71≠-7No.Hmm, none of these are working. Maybe the cubic doesn't factor nicely, and I need to use the cubic formula or numerical methods.Alternatively, perhaps I can use the rational root theorem more carefully, but since none of the possible rational roots worked, maybe the roots are irrational or complex.Wait, let me try to graph the function f(x)=x³ +6x² -7x -6 to see where it crosses zero.At x=0: f(0)=-6x=1:1 +6 -7 -6=-6x=2:8 +24 -14 -6=12So, between x=1 and x=2, the function goes from -6 to 12, so there's a root between 1 and 2.Similarly, at x=-3: -27 +54 +21 -6=42x=-4: -64 +96 +28 -6=54x=-5: -125 +150 +35 -6=54x=-6: -216 +216 +42 -6=36x=-7: -343 +294 +49 -6= -4So, between x=-7 and x=-6, f(x) goes from -4 to 36, so another root between -7 and -6.And between x=-4 and x=-3, f(x) goes from 54 to 42, so no crossing there.Wait, but that suggests two real roots, but cubic equations have three roots, so the third must be complex.But the original problem is likely expecting real solutions, so maybe there are two real roots and one complex.But the assistant's answer was x=1, -2, -3, which don't satisfy the equation, so perhaps the problem was intended to have a different setup.Alternatively, maybe I made a mistake in the cross-multiplication.Wait, let me try to solve the equation numerically.We have f(x)=x³ +6x² -7x -6We can use the Newton-Raphson method to approximate the roots.First, let's find the root between 1 and 2.Let me take x0=1.5f(1.5)=3.375 +13.5 -10.5 -6=0.375f'(x)=3x² +12x -7f'(1.5)=6.75 +18 -7=17.75Next approximation: x1=1.5 - f(x0)/f'(x0)=1.5 -0.375/17.75≈1.5 -0.0211≈1.4789f(1.4789)= (1.4789)^3 +6*(1.4789)^2 -7*(1.4789) -6≈3.227 +12.834 -10.352 -6≈-0.3Wait, that's negative, so maybe I miscalculated.Wait, let's compute f(1.4789):1.4789³≈3.2276*(1.4789)^2≈6*2.187≈13.122-7*1.4789≈-10.352-6So, total≈3.227 +13.122 -10.352 -6≈0.0Wait, that's approximately zero. So, x≈1.4789 is a root.Similarly, let's find the root between -7 and -6.Let me take x0=-6.5f(-6.5)=(-6.5)^3 +6*(-6.5)^2 -7*(-6.5) -6≈-274.625 +253.5 +45.5 -6≈18.375f'(x)=3x² +12x -7f'(-6.5)=3*(42.25) +12*(-6.5) -7≈126.75 -78 -7≈41.75Next approximation: x1=-6.5 -18.375/41.75≈-6.5 -0.44≈-6.94f(-6.94)=(-6.94)^3 +6*(-6.94)^2 -7*(-6.94) -6≈-334.0 +6*48.16 +48.58 -6≈-334 +288.96 +48.58 -6≈-334 +331.54≈-2.46f'(x)=3*(-6.94)^2 +12*(-6.94) -7≈3*48.16 -83.28 -7≈144.48 -83.28 -7≈54.2Next approximation: x2=-6.94 - (-2.46)/54.2≈-6.94 +0.045≈-6.895f(-6.895)=(-6.895)^3 +6*(-6.895)^2 -7*(-6.895) -6≈-326.0 +6*47.53 +48.265 -6≈-326 +285.18 +48.265 -6≈-326 +327.445≈1.445f'(x)=3*(-6.895)^2 +12*(-6.895) -7≈3*47.53 -82.74 -7≈142.59 -82.74 -7≈52.85Next approximation: x3=-6.895 -1.445/52.85≈-6.895 -0.027≈-6.922f(-6.922)=(-6.922)^3 +6*(-6.922)^2 -7*(-6.922) -6≈-330.0 +6*47.9 +48.454 -6≈-330 +287.4 +48.454 -6≈-330 +329.854≈-0.146f'(x)=3*(-6.922)^2 +12*(-6.922) -7≈3*47.9 + (-83.064) -7≈143.7 -83.064 -7≈53.636Next approximation: x4=-6.922 - (-0.146)/53.636≈-6.922 +0.0027≈-6.9193f(-6.9193)=(-6.9193)^3 +6*(-6.9193)^2 -7*(-6.9193) -6≈-329.0 +6*47.88 +48.435 -6≈-329 +287.28 +48.435 -6≈-329 +329.715≈0.715Wait, this is oscillating. Maybe I need a better initial guess.Alternatively, perhaps use the fact that f(-7)= -343 +294 +49 -6= -4f(-6)= -216 +216 +42 -6=36So, between x=-7 and x=-6, f(x) goes from -4 to 36, so a root exists there.Using linear approximation:The change in x is 1 (from -7 to -6), and the change in f(x) is 36 - (-4)=40.We need to find x where f(x)=0, which is 4 units above f(-7).So, fraction=4/40=0.1Thus, x≈-7 +0.1= -6.9Let me compute f(-6.9):(-6.9)^3 +6*(-6.9)^2 -7*(-6.9) -6≈-328.5 +6*47.61 +48.3 -6≈-328.5 +285.66 +48.3 -6≈-328.5 +327.96≈-0.54f(-6.8):(-6.8)^3 +6*(-6.8)^2 -7*(-6.8) -6≈-314.432 +6*46.24 +47.6 -6≈-314.432 +277.44 +47.6 -6≈-314.432 +319.04≈4.608So, f(-6.9)≈-0.54, f(-6.8)≈4.608Using linear approximation between x=-6.9 and x=-6.8:Slope=4.608 - (-0.54)=5.148 over 0.1We need to find x where f(x)=0, starting from x=-6.9, f=-0.54So, delta_x=0.54/5.148≈0.105Thus, x≈-6.9 +0.105≈-6.795Let me compute f(-6.795):(-6.795)^3 +6*(-6.795)^2 -7*(-6.795) -6≈-312.5 +6*46.18 +47.565 -6≈-312.5 +277.08 +47.565 -6≈-312.5 +318.645≈6.145Wait, that's positive, but I expected it to be near zero. Maybe my approximation was off.Alternatively, perhaps use the secant method between x=-6.9 and x=-6.8.f(-6.9)= -0.54f(-6.8)=4.608The secant method formula:x_new = x1 - f(x1)*(x1 -x0)/(f(x1)-f(x0))x0=-6.9, f(x0)=-0.54x1=-6.8, f(x1)=4.608x_new= -6.8 -4.608*(-6.8 - (-6.9))/(4.608 - (-0.54))= -6.8 -4.608*(0.1)/(5.148)= -6.8 -4.608*0.0194≈-6.8 -0.089≈-6.889Compute f(-6.889):(-6.889)^3 +6*(-6.889)^2 -7*(-6.889) -6≈-323.0 +6*47.43 +48.223 -6≈-323 +284.58 +48.223 -6≈-323 +326.803≈3.803Hmm, still positive. Maybe another iteration.x0=-6.889, f(x0)=3.803x1=-6.9, f(x1)=-0.54x_new= x1 - f(x1)*(x1 -x0)/(f(x1)-f(x0))= -6.9 - (-0.54)*(-6.9 - (-6.889))/( -0.54 -3.803)= -6.9 - (-0.54)*(-0.011)/(-4.343)= -6.9 - (0.00594)/(-4.343)≈-6.9 +0.00137≈-6.8986Compute f(-6.8986):≈-323.0 +6*47.58 +48.29 -6≈-323 +285.48 +48.29 -6≈-323 +327.77≈4.77Wait, this is getting worse. Maybe the root is near x≈-6.92.Alternatively, perhaps use a calculator or software for better approximation, but since I'm doing this manually, it's time-consuming.In any case, the cubic equation x³ +6x² -7x -6=0 has one real root near x≈1.48 and another near x≈-6.92, with the third root being complex.Therefore, the original equation has two real solutions: approximately x≈1.48 and x≈-6.92.But since the problem likely expects exact solutions, and given that the assistant's answer was incorrect due to a miscalculation, perhaps the problem was intended to have a different setup or there's a typo.Alternatively, maybe I can revisit the original equation and see if there's a different approach.Wait, perhaps instead of cross-multiplying, I can subtract the two fractions and set the result to zero:(5 + x)/(7 + x) - (2 + x²)/(4 + x) =0To combine these, find a common denominator, which is (7 + x)(4 + x):[(5 + x)(4 + x) - (2 + x²)(7 + x)] / [(7 + x)(4 + x)] =0So, the numerator must be zero:(5 + x)(4 + x) - (2 + x²)(7 + x)=0Which is the same as before, leading to the cubic equation.Therefore, unless there's a mistake in the problem statement, the solutions are not nice integers, but rather irrational numbers.Alternatively, perhaps the original problem had different coefficients, leading to a factorable cubic.In conclusion, based on the given problem, the solutions are approximately x≈1.48 and x≈-6.92, with the third root being complex. However, since the assistant's answer was x=1, -2, -3, which don't satisfy the equation, it's likely that there was an error in their solution process.</think>

🧠:<think>I need to determine which set of three line segments can form a right triangle. To do this, I'll use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.Let's examine each option one by one.Option A: 2 cm, 3 cm, 4 cm- Identify the longest side: 4 cm.- Check if (2^2 + 3^2 = 4^2): [ 4 + 9 = 13 quad text{and} quad 16 ]- Since (13 neq 16), this set does not form a right triangle.Option B: 3 cm, 4 cm, 5 cm- Identify the longest side: 5 cm.- Check if (3^2 + 4^2 = 5^2): [ 9 + 16 = 25 quad text{and} quad 25 ]- Since (25 = 25), this set forms a right triangle.Option C: 4 cm, 5 cm, 6 cm- Identify the longest side: 6 cm.- Check if (4^2 + 5^2 = 6^2): [ 16 + 25 = 41 quad text{and} quad 36 ]- Since (41 neq 36), this set does not form a right triangle.Option D: 5 cm, 6 cm, 7 cm- Identify the longest side: 7 cm.- Check if (5^2 + 6^2 = 7^2): [ 25 + 36 = 61 quad text{and} quad 49 ]- Since (61 neq 49), this set does not form a right triangle.After evaluating all options, only Option B satisfies the Pythagorean theorem and can form a right triangle.</think>

🧠:<think>Okay, so I have this problem with three equations and three unknowns: x, y, and z. All of them are positive numbers. Let me try to figure out how to solve this step by step.First, let's write down the equations again to make sure I have them right:1) x - 4 = 21/x2) x + y² = 453) y * z = x³Alright, starting with the first equation: x - 4 = 21/x. This looks like a quadratic equation. Maybe I can rearrange it to standard quadratic form. If I multiply both sides by x to eliminate the fraction, I get:x*(x - 4) = 21Which simplifies to:x² - 4x = 21Now, subtracting 21 from both sides to set it to zero:x² - 4x - 21 = 0Okay, so I have a quadratic equation here. I can try to factor it, but I'm not sure if it factors neatly. Let me check the discriminant to see if it has real roots. The discriminant D is b² - 4ac, where a = 1, b = -4, and c = -21.D = (-4)² - 4*1*(-21) = 16 + 84 = 100Since the discriminant is positive, there are two real roots. The roots can be found using the quadratic formula:x = [4 ± √100]/2 = [4 ± 10]/2So, x = (4 + 10)/2 = 14/2 = 7 or x = (4 - 10)/2 = -6/2 = -3But the problem specifies that x is a positive number, so x = 7.Great, I've found x. Now, let's move to the second equation: x + y² = 45. Since I know x is 7, I can plug that in:7 + y² = 45Subtracting 7 from both sides:y² = 38Taking the square root of both sides, since y is positive:y = √38Alright, so y is the square root of 38. Now, onto the third equation: y * z = x³. I know y and x now, so I can solve for z.First, let's compute x³:x³ = 7³ = 343So, the equation becomes:√38 * z = 343To solve for z, I'll divide both sides by √38:z = 343 / √38Hmm, that's a bit messy with the square root in the denominator. Maybe I should rationalize the denominator to make it look nicer. To do that, I'll multiply both the numerator and the denominator by √38:z = (343 * √38) / (√38 * √38) = (343√38) / 38So, z simplifies to 343√38 divided by 38.Let me just recap to make sure I didn't make any mistakes:1. Solved the first equation for x and got x = 7.2. Plugged x = 7 into the second equation to find y = √38.3. Used x = 7 and y = √38 in the third equation to find z = (343√38)/38.Everything seems to check out. I think I've got all the values correctly.</think>