Question

(x² -1)\(\left(\frac{1}{x-1}-\frac{1}{x+1}-1\right)\)

Mấy bạn giúp tui bài này với , tui sắp nộp đề cương rồi

 

Answer 1

Mk sai từ dòng 3 nhá -- 

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

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

Answer 2

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

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

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

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

Answer 3

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

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

\(=\left(x+1\right)-\left(x-1\right)-\left(x^2-1\right)\)

\(=3-x^2\)