Question

Rút gọn biểu thức

\(\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right).\frac{x^2-1}{5}\)

Answer 1

\(=\frac{\left(2x+1\right)\left(x+1\right)+8-\left(x-1\right)^2}{x^2-1}.\frac{x^2-1}{5}=\)

\(=\frac{2x^2+3x+1+8-x^2+2x-1}{5}=\frac{x^2+5x+8}{5}\)

Answer 2

\(\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right)\cdot\frac{x^2-1}{5}\left(x\ne\pm1\right)\)

\(=\left(\frac{2x+1}{x-1}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x+1}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\left(\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\left(\frac{2x^2+3x+1}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\frac{2x^2+3x+1+8-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\frac{\left(x^2+5x+8\right)\cdot\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)5}=\frac{x^2+5x+8}{5}\)

Answer 3

\(\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right).\frac{x^2-1}{5}\)

\(=\left(\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x+1\right)\left(x-1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right).\frac{\left(x+1\right)\left(x-1\right)}{5}\)

\(=\left(\frac{2x^2+3x+1}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x+1\right)\left(x-1\right)}-\frac{x^2-2x+1}{\left(x+1\right)\left(x-1\right)}\right).\frac{\left(x+1\right)\left(x-1\right)}{5}\)

\(=\left(\frac{2x^2+3x+1+8-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x+1\right)\left(x-1\right)}{5}\)

\(=\left(\frac{x^2+5x+8}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x+1\right)\left(x-1\right)}{5}=\frac{x^2+5x+8}{5}\)

Answer 4

\(\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right)\cdot\frac{x^2-1}{5}\)

\(=\frac{\left(2x+1\right)\left(x+1\right)+8-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x+1\right)\left(x-1\right)}{5}\)

\(=\frac{x^2+3x+1+8-x^2+2x-1}{5}\)

\(=\frac{5x+8}{5}\)

Answer 5

\(\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right)\cdot\frac{x^2-1}{5}\)

\(=\frac{\left(2x+1\right)\left(x+1\right)+8-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x+1\right)\left(x-1\right)}{5}\)

\(=\frac{x^2+3x+1+8-x^2+2x-1}{5}\)

\(=\frac{5x+8}{5}\)

Answer 6

Bài làm:

đkxđ: \(x\ne\pm1\)

Ta có: \(\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right).\frac{x^2-1}{5}\)

\(=\left[\frac{2x+1}{x-1}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x+1}\right].\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\frac{\left(2x+1\right)\left(x+1\right)+8-\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\frac{2x^2+3x+1+8-x^2+2x-1}{5}\)

\(=\frac{x^2+5x+8}{5}\)

Answer 7

                                                                Bài giải

\(\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right)\cdot\frac{x^2-1}{5}\)

\(=\frac{\left(2x+1\right)\left(x+1\right)+8-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x^2-1}{5}\)

\(=\frac{2x^2+3x+1+8-\left(x^2-2x+1\right)}{5}=\frac{2x^2+3x+1+8-x^2+2x-1}{5}=\frac{x^2+5x+8}{5}\)