Question

C=\(\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x+1}{x^2-1}\right)\frac{x+2006}{x}\)

a,Rút gọn

b.Tìm x∈Z để C∈Z

Answer 1

ĐKXĐ : \(x\ne0;1;-1\)

Ta có :

\(C=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x+1}{x^2-1}\right).\frac{x+2006}{x}\)

\(=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{x^2-4x+1}{\left(x-1\right)\left(x+1\right)}\right).\frac{x+2006}{x}\)

\(=\frac{\left(x+1-x+1\right)\left(x+1+x-1\right)+x^2-4x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2006}{x}\)

\(=\frac{2x+x^2-4x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2006}{x}\)

\(=\frac{\left(x-1\right)^2\left(x+2006\right)}{\left(x-1\right)\left(x+1\right).x}\)

\(=\frac{\left(x-1\right)\left(x+2006\right)}{x\left(x-1\right)}\)

Vậy....