Question

Answer 1

\(b,VP=\dfrac{x\cdot x\left(x+2\right)}{\left(x+2\right)\cdot x\left(x+2\right)}=\dfrac{x^2\left(x+2\right)}{x\left(x+2\right)^2}=VT\\ c,VP=\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}=\dfrac{3-x}{3+x}=VT\\ d,VT=\dfrac{x\left(x^2-4\right)}{5\left(2-x\right)}=\dfrac{x\left(x-2\right)\left(x+2\right)}{-5\left(x-2\right)}=\dfrac{-x\left(x+2\right)}{5}=\dfrac{-x^2-2x}{5}=VP\)

Answer 2

c: \(\dfrac{x^2-6x+9}{9-x^2}=\dfrac{\left(x-3\right)^2}{\left(3+x\right)\left(3-x\right)}=\dfrac{3-x}{3+x}\)