Question

[Mai Tuấn Anh]

cho bt \(A=\frac{x^2}{x^2-4}-\frac{x}{x+2}-\frac{2}{x-2}\)

a, tim x de ieu thuc A duoc xac dinh

b,rut gon bieu thuc A 

c,tim gia tri nguyen cua x de A nguỵen

Answer 1

[nguyen ngoc diem quynh]

dkxd  \(\hept{\begin{cases}\\\end{cases}}x-2=0;x+2=0\Leftrightarrow\hept{\begin{cases}\\\end{cases}x=+2;x=-2}\)

b/ \(\frac{x^2}{x^2-4}-\frac{x}{x+2}-\frac{2}{x-2}=\frac{x^2}{\left(x-2\right).\left(x+2\right)}-\frac{x.\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}-\frac{2.\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}\)

\(\frac{x^2-x^2-2x-2x+4}{\left(x-2\right).\left(x+2\right)}=\frac{4}{\left(x-2\right)\left(x+2\right)}\)

tới khúc này bí rồi ^^

Answer 2

[luyen hong dung]

a,ĐKXĐ của A là:\(x\ne+2;-2\)

b,\(\frac{x^2-x^2+2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{4}{\left(x+2\right)\left(x-2\right)}\)

c,Để A\(\in\)Z=> (x+2)(x-2)\(\inƯ\)(4) hay \(x^2-4\inƯ\)(4)=\(\left(4;-4;2;-2;1;-1\right)\)

Ta có bảng

\(x^2-4\)	x
4	\(\sqrt{8}\)
-4	0
2	\(\sqrt{6}\)
-2	\(\sqrt{2}\)
1	\(\sqrt{5}\)

Vậy A\(Z=>x\in\)( 0;\(\sqrt{8};\sqrt{6};\sqrt{2};\sqrt{5}\))