a, Với x khác 1 ; -1
\(G=\left(\dfrac{\left(x+1\right)^2+x\left(x-1\right)-x}{\left(x+1\right)\left(x-1\right)}\right):\left(\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right)\)
\(=\left(\dfrac{x^2+2x+1+x^2-x-x}{\left(x+1\right)\left(x-1\right)}\right):\left(\dfrac{\left(x+1-x+1\right)\left(x+1+x-1\right)}{\left(x+1\right)\left(x-1\right)}\right)\)
\(=\left(\dfrac{2x^2+1}{\left(x+1\right)\left(x-1\right)}\right):\left(\dfrac{4x}{\left(x+1\right)\left(x-1\right)}\right)=\dfrac{2x^2+1}{4x}\)
b, Ta có \(\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Với x = 5 thì G = \(\dfrac{2.25+1}{20}=\dfrac{51}{20}\)
Với x = 1 thì G = \(\dfrac{2+1}{4}=\dfrac{3}{4}\)
