Question

Everyone please help me this in the photo (⁠ᗒ⁠ᗩ⁠ᗕ⁠)​

Answer 1

`\textsf{1.} \quad (20xy^2)/(7(xy-3))`

`\textsf{2.} \quad (1)/((x+5)(x-3))`

`\textsf{3.} \quad (a-1)/((a+1)(a-2))}`

Explanation:

Question 1

Given:

`\left((5(y+6))/(7(xy-3))\right) * \left((4xy^2)/(y+6)\right)`

`\textsf{Apply\:the\:fraction\:rule}:\quad (a)/(b)* (c)/(d)=(a* c)/(b* d)`

`\implies(5(y+6)4xy^2)/(7(xy-3)(y+6))`

`\textsf{Multiply the $5$ and $4$}:`

`\implies (20xy^2(y+6))/(7(xy-3)(y+6))`

`\textsf{Cancel the common factor $(y + 6)$}:`

`\implies (20xy^2)/(7(xy-3))`

Question 2

Given:

`\left((4)/(x^2-25)\right) * \left((x-5)/(4x-12)\right)`

`\textsf{Apply\:the\:fraction\:rule}:\quad (a)/(b)* (c)/(d)=(a* c)/(b* d)`

`\implies (4(x-5))/((x^2-25)(4x-12))`

`\textsf{Factor $x^2-25: \quad (x+5)(x-5)$}`

`\textsf{Factor $4x-12: \quad 4(x-3)$}`

`\implies (4(x-5))/(4(x+5)(x-5)(x-3))`

`\textsf{Cancel the common factor $4(x - 5)$}:`

`\implies (1)/((x+5)(x-3))`

Question 3

Given:

`\left((a^2-1)/(a^3+1)}\right) * \left((a^2-a+1)/(a^2-a-2)\right)`

`\textsf{Apply\:the\:fraction\:rule}:\quad (a)/(b)* (c)/(d)=(a* c)/(b* d)`

`\implies ((a^2-1)(a^2-a+1))/((a^3+1)(a^2-a-2))`

`\textsf{Factor $a^2-1: \quad (a+1)(a-1)$}`

`\textsf{Factor $a^3+1: \quad (a+1)(a^2-a+1)$}`

`\implies ((a+1)(a-1)(a^2-a+1))/((a+1)(a^2-a+1)(a^2-a-2))}`

`\textsf{Cancel the common factor $(a + 1)(a^2- a + 1)$}:`

`\implies (a-1)/(a^2-a-2)`

`\textsf{Factor the denominator}:`

`\implies (a-1)/((a+1)(a-2))}`

Formulas used:

`\boxed{\begin{minipage}{5.5cm}\underline{Difference of Two Squares Formula}\\\\$x^2-y^2=(x+y)(x-y)$\\\\ \underline{Sum of Cubes Formula}\\\\$x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)$\\\end{minipage}}`