Question

Cho 2 biểu thức : \(A=\left(\frac{x}{x+4}+\frac{4}{x-4}\right):\frac{x^2+16}{x+2}\&B=\frac{x+4}{x+2}\left(x>0,x#4\right)\)

a, CMR : \(A=\frac{x+2}{x^2-16}\)

b, Tìm \(x\in Z\)để \(M=A\left(B-1\right)\in Z\)

Answer 1

â) \(A=\left(\frac{x}{x+4}+\frac{4}{x-4}\right):\frac{x^2+16}{x+2}\) 

\(=\left(\frac{x\left(x-4\right)+4\left(x+4\right)}{\left(x+4\right)\left(x-4\right)}\right)=\left(\frac{x^2+16}{x^2-16}\right):\frac{x^2+16}{x+2}\)  

\(=\frac{x+2}{x^2-16}\left(đpcm\right)\)

Answer 2

a) \(A=\left(\frac{x}{x+4}+\frac{4}{x-4}\right):\frac{x^2+16}{x+2}\)

\(A=\frac{x\left(x-4\right)+4\left(x+4\right)}{\left(x+4\right)\left(x-4\right)}.\frac{x+2}{x^2+16}\)

\(A=\frac{x^2-4x+4x+16}{x^2-16}.\frac{x+2}{x^2+16}\)

\(A=\frac{x^2+16}{x^2-16}.\frac{x+2}{x^2+16}\)

\(A=\frac{x+2}{x^2-16}\left(đpcm\right)\)