Rút gọn: A = \(\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
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30 tháng 9 2016
\(\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2-\frac{x^2-10}{x+2}\right)\left(ĐK:x\ne\pm2\right)\)
\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{\left(x-2\right)\left(x+2\right)-\left(x^2-10\right)}{x+2}\)
\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{x^2-4-x^2+10}\)
\(=\frac{-6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}=-\frac{1}{x-2}=\frac{1}{2-x}\)
20 tháng 12 2016
\(A=\left(\frac{x-1}{x-2}+\frac{x+3}{x^2-4}\right):\left(\frac{x+2}{x-2}+\frac{1}{2-x}\right)\)
\(A=\frac{\left(x-1\right)\left(x+2\right)+x+3}{\left(x+2\right)\left(x-2\right)}:\left(\frac{x+2}{x-2}-\frac{1}{x-2}\right)\)
\(A=\frac{x^2+2x-x-2+x+3}{\left(x+2\right)\left(x-2\right)}:\frac{x+2-1}{x-2}\)
\(A=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}.\frac{x-2}{x+1}\)
\(A=\frac{\left(x+1\right)^2}{x+2}.\frac{1}{x+1}\)
\(A=\frac{x+1}{x+2}\)
30 tháng 1 2022
\(A=\left(\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\cdot\dfrac{2-x}{x}\)
\(=\dfrac{x+2+2x+x-2}{-\left(2-x\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}\)
\(=\dfrac{4x}{-\left(x+2\right)\cdot x}=\dfrac{-4}{x+2}\)
30 tháng 3 2021
a) ĐKXĐ : x ≠ ±2
\(=\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\div\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(=\left[\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right]\div\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\div\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\times\frac{x+2}{6}=\frac{-1}{x-2}\)
b) Để A < 0 thì -1/x-2 < 0
=> x - 2 > 0 <=> x > 2
Vậy với x > 2 thì A < 0
ĐKXĐ : \(x\ne\pm2\)
\(A=\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\div\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(=\left[\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right]\div\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\times\frac{x+2}{6}=-\frac{1}{x-2}\)