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For each expression, simplify if possible. If applicable, click on "Cannot be simplified". \( \frac{\begin{array}{l}\frac{6 v+5}{5 v+6} \\ \frac{v-7}{-7+v}\end{array}=\square}{\square} \)

Ask by Hodgson West. in the United States
Feb 26,2025

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Tutor-Verified Answer

Answer

The expression simplifies to \( \frac{6v+5}{5v+6} \).

Solution

Calculate or simplify the expression \( \frac{\frac{6v+5}{5v+6}}{\frac{v-7}{-7+v}} \). Simplify the expression by following steps: - step0: Solution: \(\frac{\frac{6v+5}{5v+6}}{\frac{v-7}{-7+v}}\) - step1: Reduce the fraction: \(\frac{\frac{6v+5}{5v+6}}{1}\) - step2: Divide the terms: \(\frac{6v+5}{5v+6}\) The expression \( \frac{\frac{6v+5}{5v+6}}{\frac{v-7}{-7+v}} \) simplifies to \( \frac{6v+5}{5v+6} \).

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Beyond the Answer

Let's simplify the given expression step by step: 1. The first part of the expression is \( \frac{6v + 5}{5v + 6} \). This fraction does not simplify further in an obvious way, as there are no common factors. 2. The second part is \( \frac{v - 7}{-7 + v} \). Notice that \( -7 + v \) can be rewritten as \( v - 7 \). Therefore, \( \frac{v - 7}{-7 + v} = \frac{v - 7}{v - 7} = 1 \) (for \( v \neq 7 \)). Putting it together, the entire expression simplifies to: \[ \frac{6v + 5}{5v + 6} = \frac{6v + 5}{5v + 6} \quad \text{and} \quad \frac{v - 7}{-7 + v} = 1 \] The final result is: \[ \frac{6v + 5}{5v + 6} = 1 \implies \text{Cannot be simplified further.} \]

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ISCELÁNEA cribir, por simple inspección, el resultado de: \( \begin{array}{lll}(x+2)^{2} & \text { 14. }(x+y+1)(x-y-1) & \text { 27. }\left(2 a^{3}-5 b^{4}\right)^{2} \\ (x+2)(x+3) & \text { 15. }(1-a)(a+1) & \text { 28. }\left(a^{3}+12\right)\left(a^{3}-15\right) \\ (x+1)(x-1) & \text { 16. }(m-8)(m+12) & \text { 29. }\left(m^{2}-m+n\right)\left(n+m+m^{2}\right) \\ (x-1)^{2} & \text { 17. }\left(x^{2}-1\right)\left(x^{2}+3\right) & \text { 30. }\left(x^{4}+7\right)\left(x^{4}-11\right) \\ (n+3)(n+5) & \text { 18. }\left(x^{3}+6\right)\left(x^{3}-8\right) & \text { 31. }(11-a b)^{2} \\ (m-3)(m+3) & \text { 19. }\left(5 x^{3}+6 m^{4}\right)^{2} & \text { 32. }\left(x^{2} y^{3}-8\right)\left(x^{2} y^{3}+6\right) \\ (a+b-1)(a+b+1) & \text { 20. }\left(x^{4}-2\right)\left(x^{4}+5\right) & \text { 33. }(a+b)(a-b)\left(a^{2}-b^{2}\right) \\ (1+b)^{3} & \text { 21. }(1-a+b)(b-a-1) & \text { 34. }(x+1)(x-1)\left(x^{2}-2\right) \\ \left(a^{2}+4\right)\left(a^{2}-4\right) & \text { 22. }\left(a^{x}+b^{n}\right)\left(a^{x}-b^{n}\right) & \text { 35. }(a+3)\left(a^{2}+9\right)(a-3) \\ \left(3 a b-5 x^{2}\right)^{2} & \text { 23. }\left(x^{a+1}-8\right)\left(x^{a+1}+9\right) & \text { 36. }(x+5)(x-5)\left(x^{2}+1\right) \\ (a b+3)(3-a b) & \text { 24. }\left(a^{2} b^{2}+c^{2}\right)\left(a^{2} b^{2}-c^{2}\right) & \text { 37. }(a+1)(a-1)(a+2)(a-2) \\ (1-4 a x)^{2} & \text { 25. }(2 a+x)^{3} & \text { 36. }\left(x^{2}-11\right)\left(x^{2}-2\right)\end{array} \) \( \left(a^{2}+8\right)\left(a^{2}-7\right) \)
Algebra Mexico Feb 26, 2025
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