\(A=\left(\dfrac{x}{x+1}+1\right):\left(1-\dfrac{3x^2}{1-x^2}\right)\)
a) Để biểu thức A được xác định thì :
\(\left\{{}\begin{matrix}x+1\ne0\\1-x^2\ne0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x\ne-1\\x\ne1\end{matrix}\right.\)
\(A=\left(\dfrac{x}{x+1}+1\right):\left(1-\dfrac{3x^2}{1-x^2}\right)\)
\(A=\left(\dfrac{x}{x+1}+\dfrac{x+1}{x+1}\right):\left(\dfrac{1-x^2}{1-x^2}-\dfrac{3x^2}{1-x^2}\right)\)
\(A=\left(\dfrac{x+x+1}{x+1}\right):\left(\dfrac{1-x^2-3x^2}{1-x^2}\right)\)
\(A=\dfrac{2x+1}{x+1}:\dfrac{\left(1-2x\right)\left(1+2x\right)}{\left(1-x\right)\left(1+x\right)}\)
\(A=\dfrac{2x+1}{x+1}.\dfrac{\left(x+1\right)\left(1-x\right)}{\left(2x+1\right)\left(1-2x\right)}\)
\(A=\dfrac{1-x}{1-2x}\)
b) Thay vào biểu thức x=2 ta có :
\(A=\dfrac{1-2}{1-2.2}=\dfrac{-1}{-3}=\dfrac{1}{3}\)
