Question

Answer 1

\(VT=\dfrac{\left(x^2+2\right)^2-4x^2}{y\left(x^2+2\right)-2xy-x^2+2x-1-1}\)

         \(=\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{y\left(x^2+2-2x\right)-\left(x^2-2x+2\right)}\)

         \(=\dfrac{\left(x^2-2x+2\right)\left(x^2+2x+2\right)}{\left(y-1\right)\left(x^2-2x+2\right)}=\dfrac{x^2+2x+2}{y-1}\)

\(\Rightarrow VT=VP\)

Answer 2

\(VT=\dfrac{3n-2-m\left(3n-2\right)}{1-3.m.1+3.m^2.1-m^3}=\dfrac{\left(3n-2\right)\left(1-m\right)}{\left(1-m\right)^3}\)

  \(=\dfrac{3n-2}{\left(1-m\right)^2}\Rightarrow VT=VP\)