Question

rút gọn và tìm x nguyên để Q có giá trị nguyên  

Answer 1

\(Q=1+\left(\dfrac{x+1}{\left(x+1\right)\cdot\left(x^2-x+1\right)}+\dfrac{1}{x^2-x+1}-\dfrac{2}{x+1}\right)\cdot\dfrac{x\left(x^2-x-1\right)}{x^2\left(x-2\right)}\)

\(=1+\dfrac{x+1+x+1-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\dfrac{x^2-x-1}{x\left(x-2\right)}\)

\(=1+\dfrac{-2x^2+4x}{\left(x+1\right)}\cdot\dfrac{1}{x\left(x-2\right)}\)

\(=1+\dfrac{-2x\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}=1-\dfrac{2}{x+1}=\dfrac{x+1-2}{x+1}=\dfrac{x-1}{x+1}\)

Để Q nguyên thì x+1-2 chia hêt cho x+1

=>\(x+1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{-2;-3\right\}\)