Question

A= \(\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right):\frac{5}{4x^2-4}\)

Tìm Tập xác định 

Rút gọn A

Answer 1

\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right):\frac{4}{4x^2-4}\)

\(=\left(\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+2\right)}+\frac{6}{2.\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\frac{4}{4\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}.\left(x-1\right)\left(x+1\right)=\frac{4}{2}=2\)

Answer 2

thêm ĐK: x khác 1 ; -1

Answer 3

tập xđ: x khác (-1,1)

A=(\(\frac{-x^2-2x-6-x^2-2x+3}{2\left(1-x^2\right)}\):\(\frac{5}{4\left(1-x^2\right)}\)

A=\(\frac{-4x^2-8x-6}{5}\)

Answer 4

bn Đặng Minh Triều ơi, phải là \(\frac{5}{4x^2-4}\) chứ