Question

\(\frac{x+1}{x-1}+\frac{x-2}{x+1}-\frac{2x^2+x+5}{x^2-1}\) 

a) Tìm điều kiện xác định của B

b) Rút gọn B

c) Tìm x để B < 2

Answer 1

a) ĐKXĐ: x khác +-1

b) \(\frac{x+1}{x-1}+\frac{x-2}{x+1}-\frac{2x^2+x+5}{x^2-1}\)

\(=\frac{x+1}{x-1}+\frac{x-2}{x+1}-\frac{2x^2+x+5}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x-2\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x^2+x+5}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{\left(x+1\right)^2+\left(x-2\right)\left(x-1\right)-\left(2x^2+x+5\right)}{\left(x-1\right)\left(x+1\right)}\)

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

\(=-\frac{2}{x-1}\)