\(A=\frac{2x}{x+1}+\frac{-x+1}{x}\)
a: Điều kiện xác định
\(\Rightarrow\orbr{\begin{cases}x+1\ne0\\x\ne0\end{cases}}\Rightarrow\orbr{\begin{cases}x\ne-1\\x\ne0\end{cases}}\)
b, \(A=\frac{2x}{x+1}+\frac{-x+1}{x}\)
\(A=\frac{2x^2}{x\left(x+1\right)}+\frac{\left(x+1\right)\left(1-x\right)}{x\left(x+1\right)}\)
\(A=\frac{2x^2+x-x^2+1-x}{x\left(x+1\right)}\)
\(A=\frac{x^2+1}{x\left(x+1\right)}\)
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