Question

Tìm x để A nhận giá trị nguyên

A=\(\dfrac{x}{x+2}\) +\(\dfrac{2}{x-2}\) +\(\dfrac{2x+4}{4-x^2}\) 

Giúp em với ạ 

Em cần gấp

 

Answer 1

\(A=\dfrac{x}{x+2}+\dfrac{2}{x-2}+\dfrac{2x+4}{4-x^2}\left(ĐKXĐ:x\ne\pm2\right)\)

\(A=\dfrac{x}{x+2}+\dfrac{2}{x-2}-\dfrac{2x+4}{x^2-4}\)

\(A=\dfrac{x}{x+2}+\dfrac{2}{x-2}-\dfrac{2x+4}{\left(x-2\right)\left(x+2\right)}\)

\(A=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x+4}{\left(x-2\right)\left(x+2\right)}\)

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

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

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

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

\(A=\dfrac{x}{x+2}\)

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

\(A=\dfrac{x+2}{x+2}+\dfrac{1}{x+2}\)

\(A=1+\dfrac{1}{x+2}\)

\(MàA\in Z\)

\(\Rightarrow\dfrac{1}{x+2}\in Z\)

\(\Rightarrow x+2\inƯ\left(1\right)=\left\{\pm1\right\}\)

Xét bảng:

\(x+2\)	\(1\)	\(-1\)
\(x\)	\(-1\left(tm\right)\)	\(-3\left(tm\right)\)