\(\text{a) }A=\dfrac{x}{2x+2}+\dfrac{x^2+1}{2-2x^2}\\ A=\dfrac{x}{2\left(x+1\right)}+\dfrac{x^2+1}{2\left(1-x^2\right)}\\ A=\dfrac{x}{2\left(x+1\right)}+\dfrac{x^2+1}{2\left(1-x\right)\left(1+x\right)}\\ A=\dfrac{x\left(1-x\right)}{2\left(x+1\right)\left(1-x\right)}+\dfrac{x^2+1}{2\left(x+1\right)\left(1-x\right)}\\ A=\dfrac{x-x^2+x^2+1}{2\left(x+1\right)\left(1-x\right)}\)
\(\Rightarrow\) Để \(A\) có nghĩa
\(\text{thì }\Rightarrow2\left(x+1\right)\left(1-x\right)\ne0\\ \Rightarrow\left\{{}\begin{matrix}x+1\ne0\\1-x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne1\end{matrix}\right.\)
\(\text{b) }A=\dfrac{x-x^2+x^2+1}{2\left(x+1\right)\left(1-x\right)}\\ A=\dfrac{x+1}{2\left(x+1\right)\left(1-x\right)}\\ A=\dfrac{1}{2\left(1-x\right)}\\ A=\dfrac{1}{2-2x}\)
c) Để \(A=\dfrac{1}{2}\)
\(\text{thì }\Rightarrow\dfrac{1}{2-2x}=\dfrac{1}{2}\\ \Leftrightarrow2-2x=2\\ \Leftrightarrow2x=0\\ \Leftrightarrow x=0\)
Vậy......................
