Bài 3:
a: \(A=\dfrac{x^2+x-2-x^2+x-2}{\left(x+1\right)^2\left(x-1\right)}\cdot\dfrac{\left(x+1\right)^2\left(x-1\right)}{x\left(2x+1\right)}\)
\(=\dfrac{2}{2x+1}\)
Bài 3:
\(a,ĐK:x\ne\pm1;x\ne0;x\ne-\dfrac{1}{2}\\ A=\dfrac{x^2+x-2-x^2+x+2}{\left(x+1\right)^2\left(x-1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)^2}{x\left(2x+1\right)}\\ A=\dfrac{2x}{x\left(2x+1\right)}=\dfrac{2}{2x+1}\\ b,x=-3\Leftrightarrow A=\dfrac{2}{-6+1}=-\dfrac{2}{5}\\ x=\dfrac{1}{4}\Leftrightarrow A=\dfrac{2}{\dfrac{1}{2}+1}=\dfrac{4}{3}\\ x=-\dfrac{1}{2}\Leftrightarrow A\in\varnothing\)
\(c,A=3\Leftrightarrow2x+1=\dfrac{2}{3}\Leftrightarrow x=-\dfrac{1}{6}\\ d,A=\dfrac{2}{3}\Leftrightarrow2x+1=3\Leftrightarrow x=1\)