Question

Cho \(A=\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{x^2-4}\). Rút gọn A

Answer 1

chịu ko bt *_*

Answer 2

ĐKXĐ : x\(\ne\mp2\)

A = \(\frac{x}{x-2}\)+\(\frac{2-x}{x+2}\)+\(\frac{12-10x}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)+\(\frac{\left(2-x\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)+\(\frac{12-10x}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{x^2+2x-x^2+4x-4+12-10x}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{8-4x}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{-4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

=\(\frac{-4}{x+2}\)

Answer 3

\(ĐKXĐ:x\ne\pm2\)

\(A=\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{x^2-4}\)\(=\frac{x}{x-2}+\frac{-\left(x-2\right)}{x+2}+\frac{-\left(10x-12\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}+\frac{-10x+12}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+2x}{\left(x-2\right)\left(x+2\right)}+\frac{-x^2+4x-4}{\left(x-2\right)\left(x+2\right)}+\frac{-10x+12}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+2x-x^2+4x-4-10x+12}{\left(x-2\right)\left(x+2\right)}=\frac{-4x+8}{\left(x-2\right)\left(x+2\right)}=\frac{-4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{-4}{x+2}\)

Answer 4

ĐKXĐ: \(x\ne\pm2\)

\(A=\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{x^2-4}\)

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

\(A=\frac{4\left(-x+2\right)}{\left(x+2\right)\left(x-2\right)}\)

\(A=\frac{-4}{x+2}\)