Ta có: \(B=\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\)
\(=\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\cdot\left(x-1\right)}\)
\(=\dfrac{x^2-1-x^2+x+2}{x\left(x-1\right)}=\dfrac{x+1}{x\left(x-1\right)}\)
M=B:A
\(=\dfrac{x+1}{x\left(x-1\right)}:\dfrac{x}{x-1}\)
\(=\dfrac{x+1}{x\left(x-1\right)}\cdot\dfrac{x-1}{x}=\dfrac{x+1}{x^2}\)
\(N=M\left(x-2\right)=\dfrac{x+1}{x^2}\cdot\left(x-2\right)\)
\(=\dfrac{x^2-x-2}{x^2}=1-\dfrac{1}{x}-\dfrac{2}{x^2}\)
\(=-2\left(\dfrac{1}{x^2}+\dfrac{1}{2x}-\dfrac{1}{2}\right)\)
\(=-2\left(\dfrac{1}{x^2}+2\cdot\dfrac{1}{x}\cdot\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{9}{16}\right)\)
\(=-2\left(\dfrac{1}{x}+\dfrac{1}{4}\right)^2+\dfrac{9}{8}< =\dfrac{9}{8}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=-4
