chịu ko bt *_*

ĐKXĐ : x\(\ne\mp2\)

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

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

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

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

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

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

\(ĐKXĐ:x\ne\pm2\)

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

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

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

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

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

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

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

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

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

b) ĐKXĐ: x \(\ne\) \(\pm\)2

Để A \(\in\)Z <=> \(-\frac{4}{x+2}\in Z\) <=> -4 \(⋮\)x + 2

<=> x + 2 \(\in\)Ư(-4) = {1; -1; 2; -2; 4; -4}

Lập bảng :

a) Rút gọn:

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

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

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

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

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

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

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

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

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

Chúc bạn học tốt!

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

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

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

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

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

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

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

b)(ĐKXĐ của A là x\(\ne\pm2\))

Với x\(\ne\pm2\) ta có:

A\(\in Z\)

\(\Leftrightarrow\frac{-4}{x+2}\in Z\)

\(\Rightarrow x+2\inƯ_{\left(-4\right)}=\left\{\pm1;\pm2;\pm4\right\}\)

Ta có bảng sau :

Vậy để \(A\in Z\) thì x = {-6,-4,-3,-1,0}

\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\frac{x^2+12}{4-x^2}\)                    ĐKXĐ: \(x\ne\pm2\)

\(=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)

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

\(=\frac{x^2-7x+10}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x^2-2x-5x+10}{\left(x+2\right)\left(x-2\right)}\)

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

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

\(=\frac{x-5}{x+2}\)

a: ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)

b: \(P=\left(\dfrac{x}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right):\left(\dfrac{10x-25}{x\left(x+5\right)}-\dfrac{x}{x-5}\right)\)

\(=\dfrac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}:\dfrac{\left(10x-25\right)\left(x-5\right)-x^2\left(x+5\right)}{x\left(x+5\right)\left(x-5\right)}\)

\(=\dfrac{10x-25}{10x^2-50x-25x+125-x^3-5x^2}\)

\(=\dfrac{10x-25}{-x^3+5x^2-75x+125}\)

\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\frac{x^2+12}{4-x^2}=\frac{\left(x-1\right).\left(x-2\right)}{x^2-4}-\frac{\left(x+2\right)^2}{x^2-4}+\frac{x^2+12}{x^2-4}\)

  \(=\frac{x^2-3x+2}{x^2-4}-\frac{x^2+4x+4}{x^2-4}+\frac{x^2+12}{x^2-4}=\frac{x^2-7x+10}{x^2-4}=\frac{\left(x-2\right).\left(x-5\right)}{\left(x-2\right).\left(x+2\right)}=\frac{x-5}{x+2}\)

\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\)\(\frac{x^2+12}{4-x^2}\)\(ĐKXĐ\): \(x\ne\pm2\)

\(=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)\(-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)\(+\frac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)

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

\(=\frac{x^2-7x+10}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x^2-2x-5x+10}{\left(x+2\right)\left(x-2\right)}\)

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

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

\(=\frac{x-5}{x+2}\)