a: ĐKXĐ: \(x\notin\left\{-1;0\right\}\)
b: \(A=\dfrac{2x^2+x^2-1-3x^2+2x+1}{x\left(x+1\right)}=\dfrac{2x}{x\left(x+1\right)}=\dfrac{2}{x+1}\)
e: Để A nguyên thì \(x+1\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{-2;1;-3\right\}\)
\(a.ĐKXĐ:\left\{{}\begin{matrix}x\ne-1\\x\ne0\end{matrix}\right.\)
\(b.A=\dfrac{2x}{x+1}+\dfrac{x-1}{x}-\dfrac{3x^2-2x-1}{x^2+x}\)
\(A=\dfrac{2x^2}{x\left(x+1\right)}+\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x+1\right)}-\dfrac{3x^2-2x-1}{x\left(x+1\right)}\)
\(A=\dfrac{2x^2+x^2-1-3x^2+2x+1}{x\left(x+1\right)}\)
\(A=\dfrac{2x}{x\left(x+1\right)}\)
\(A=\dfrac{2}{x+1}\)
\(c.\dfrac{2}{x+1}=\dfrac{1}{2}\)
\(\Leftrightarrow x+1=4\)
\(\Leftrightarrow x=4-1=3\left(thỏa.mãn\right)\)