mk nghỉ bài này đề sai
a) điều kiện : \(x\ne0;x\ne-1;x\ne2\)
ta có : \(A=1+\left(\dfrac{x+1}{x^3+1}-\dfrac{1}{x-x^2-1}+\dfrac{2}{x+1}\right):\dfrac{x^3-2x^2}{x^3-x^2+x}\)
\(\Leftrightarrow A=1+\left(\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x^2-x+1}+\dfrac{2}{x+1}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{x+1+x+1+2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{2x^2+4}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x^2-x+1}{x\left(x-2\right)}\) \(\Leftrightarrow A=1+\dfrac{2x^2+4}{x\left(x+1\right)\left(x-2\right)}=\dfrac{2x^2+4+x\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}\)\(\Leftrightarrow A=\dfrac{x^3+x^2-2x+4}{x\left(x+1\right)\left(x-2\right)}\)
b) ta có : \(\left|x-\dfrac{3}{4}\right|=\dfrac{5}{4}\) \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{5}{4}\\x-\dfrac{3}{4}=\dfrac{-5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=\dfrac{-1}{2}\end{matrix}\right.\)
thế vào \(A\) ta có : \(A=\dfrac{41}{5}\)
vậy ...............................................................................................................
