Question

[nguyen ngoc son]

cho biểu thức

\(A=\left(\dfrac{x}{x^2-4}-\dfrac{4}{2-x}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)

a.rút gọn A

b.tìm giá trị nguyên của x để A nhận giá trị nguyên

helpp

Answer 1

[Nguyễn Lê Phước Thịnh]

ĐKXĐ: \(x\notin\left\{2;-2;-1\right\}\)

a) Ta có: \(A=\left(\dfrac{x}{x^2-4}-\dfrac{4}{2-x}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}+\dfrac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right):\dfrac{3\left(x+1\right)}{x\left(x+2\right)}\)

\(=\left(\dfrac{x+4x+8}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

\(=\dfrac{5x+8+x-2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

\(=\dfrac{6x+6}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

\(=\dfrac{6\left(x+1\right)}{x-2}\cdot\dfrac{x}{3\left(x+1\right)}\)

\(=\dfrac{2x}{x-2}\)

b) Để A nguyên thì \(2x⋮x-2\)

\(\Leftrightarrow2x-4+4⋮x-2\)

mà \(2x-4⋮x-2\)

nên \(4⋮x-2\)

\(\Leftrightarrow x-2\inƯ\left(4\right)\)

\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Leftrightarrow x\in\left\{3;1;4;0;6;-2\right\}\)

Kết hợp ĐKXĐ, ta được:

\(x\in\left\{0;1;3;4;6\right\}\)

Vậy: Khi \(x\in\left\{0;1;3;4;6\right\}\) thì A nguyên