Question

rút gọn chi tiết

bài 2

\(P=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right).\dfrac{x+2017}{x}\) vs \(x\ne0,x\pm1\)

a/ rút gọn P

b/ tìm giá trị nguyên của x để P có giá trị nguyên

Answer 1

a: \(P=\left(\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x+2017}{x}\)

\(=\dfrac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+2017}{x}\)

\(=\dfrac{x^2-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+2017}{x}=\dfrac{x+2017}{x}\)

b: Để P là số nguyên thì \(x+2017⋮x\)

\(\Leftrightarrow x\in\left\{2017;-2017\right\}\)