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ON
1
BP
25 tháng 12 2019
\(A=\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{x^2-4}\)
\(=\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(2-x\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12-10x}{\left(x-2\right)\left(x+2\right)}=\frac{x^2+2x+4x-4-x^2+12-10x}{\left(x-2\right)\left(x+2\right)}=\frac{8-4x}{\left(x-2\right)\left(x+2\right)}=\frac{-4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{4}{x+2}\)
4 tháng 1 2020
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 :
| x + 2 | 1 | -1 | 2 | -2 | 4 | -4 |
| x | -1 | -3 | 0 | -4 | 2(ktm) | -6 |
VM
4 tháng 1 2020
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!
L
1
17 tháng 7 2018
\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\frac{x^2+12}{4-x^2}\) ĐKXĐ: \(x\ne\pm2\)
\(=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2-2x-x+2-x^2-4x-4+x^2+12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-7x+10}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-2x-5x+10}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x\left(x-2\right)-5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-5\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-5}{x+2}\)
VV
2
3 tháng 7 2018
\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\frac{x^2+12}{4-x^2}=\frac{\left(x-1\right).\left(x-2\right)}{x^2-4}-\frac{\left(x+2\right)^2}{x^2-4}+\frac{x^2+12}{x^2-4}\)
\(=\frac{x^2-3x+2}{x^2-4}-\frac{x^2+4x+4}{x^2-4}+\frac{x^2+12}{x^2-4}=\frac{x^2-7x+10}{x^2-4}=\frac{\left(x-2\right).\left(x-5\right)}{\left(x-2\right).\left(x+2\right)}=\frac{x-5}{x+2}\)
3 tháng 7 2018
\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\)\(\frac{x^2+12}{4-x^2}\)\(ĐKXĐ\): \(x\ne\pm2\)
\(=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)\(-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)\(+\frac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2-2x-x+2-x^2-4x-4+x^2+12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-7x+10}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-2x-5x+10}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x\left(x-2\right)-5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-5\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-5}{x+2}\)
2 tháng 8 2021
\(ĐK:x\ne\pm3\)
\(P=\left[\frac{\left(2x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3-10x}{\left(x-3\right)\left(x+3\right)}\right]\cdot\frac{x-3}{x+2}\)
\(=\frac{2x^2-7x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x-3}{x+2}\)
\(=\frac{3x^2+6x}{x+3}\cdot\frac{1}{x+2}=\frac{3x\left(x+2\right)}{\left(x+3\right)\left(x+2\right)}=\frac{3x}{x+3}\)
chịu ko bt *_*
ĐKXĐ : x\(\ne\mp2\)
A = \(\frac{x}{x-2}\)+\(\frac{2-x}{x+2}\)+\(\frac{12-10x}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)+\(\frac{\left(2-x\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)+\(\frac{12-10x}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{x^2+2x-x^2+4x-4+12-10x}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{8-4x}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{-4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{-4}{x+2}\)