1: \(M=\dfrac{1}{x+1}-\dfrac{x^3-x}{x^2+1}\cdot\dfrac{1}{x^2+2x+1}-\dfrac{1}{x^2-1}\)
\(=\dfrac{1}{x+1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{\left(x^2+1\right)\left(x+1\right)^2}-\dfrac{1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x-2}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x^2+1\right)\left(x+1\right)}\)
\(=\dfrac{x^3+x-2x^2-2-x\left(x^2-1\right)}{\left(x+1\right)\left(x-1\right)\left(x^2+1\right)}\)
\(=\dfrac{x^3-2x^2+x-2-x^3+x}{\left(x+1\right)\left(x-1\right)\left(x^2+1\right)}\)
\(=\dfrac{-2x^2+2x-2}{\left(x+1\right)\left(x-1\right)\left(x^2+1\right)}\)
2: Để M=1 thì \(-2x^2+2x-2=\left(x^2-1\right)\left(x^2+1\right)\)
\(\Leftrightarrow x^4-1+2x^2-2x+2=0\)
\(\Leftrightarrow x^4+2x^2-2x+1=0\)
hay \(x\in\varnothing\)
