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20 tháng 12 2016
\(A=\left(\frac{x-1}{x-2}+\frac{x+3}{x^2-4}\right):\left(\frac{x+2}{x-2}+\frac{1}{2-x}\right)\)
\(A=\frac{\left(x-1\right)\left(x+2\right)+x+3}{\left(x+2\right)\left(x-2\right)}:\left(\frac{x+2}{x-2}-\frac{1}{x-2}\right)\)
\(A=\frac{x^2+2x-x-2+x+3}{\left(x+2\right)\left(x-2\right)}:\frac{x+2-1}{x-2}\)
\(A=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}.\frac{x-2}{x+1}\)
\(A=\frac{\left(x+1\right)^2}{x+2}.\frac{1}{x+1}\)
\(A=\frac{x+1}{x+2}\)
TT
3
21 tháng 10 2018
\(ĐKXĐ:\hept{\begin{cases}x-2\ne0\\x^2-2x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\left(x-2\right)\ne0\Leftrightarrow x\ne0\end{cases}.}}\)
\(A=\left(\frac{x}{x-2}+\frac{2x}{x^2-2x}\right)\left(x^2+4\right)\)
\(=\left(\frac{x}{x-2}+\frac{2x}{x\left(x-2\right)}\right)\left(x^2+4\right)\)
\(=\frac{x+2}{x-2}.\left(x^2+4\right)\)(x^2-4 còn rút gọn đc thế này thì bó tay)
21 tháng 10 2018
( Sai dấu )
ĐKXĐ
\(\hept{\begin{cases}x-2\ne0\\x\left(x-2\right)\ne0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x\ne2\\x\ne0\end{cases}}\) ( T/m đk )
\(A=\left(\frac{x}{x-2}+\frac{2x}{x^2-2x}\right).\left(x^2-4\right)\)
\(=\left[\frac{x}{x-2}+\frac{2x}{x\left(x-2\right)}\right]\left(x^2-4\right)\)
\(=\left[\frac{x}{x-2}+\frac{2}{\left(x-2\right)}\right]\left(x^2-4\right)\)
\(=\frac{x+2}{x-2}.\left(x^2-4\right)\)
\(=\frac{x+2.x^2+4}{x+2}=\frac{x+2}{x-2}.\left(x-2\right)\left(x+2\right)\)
\(=\frac{x+2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)}=\left(x+2\right)^2\)
13 tháng 3 2020
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(ĐKXĐ:x\ne\pm2\)
\(A=\left(\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x}{\left(x+2\right)^2}\frac{\left(x-2\right)\left(x+2\right)}{-x}\)
\(=\frac{-2\left(x-2\right)}{x+2}=\frac{4-2x}{x+2}\)
HH
13 tháng 3 2020
\(ĐKXĐ:x\ne\pm2;x\ne0\)
\(A=\left(\frac{2}{2+x}-\frac{4}{x^2+4x +4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(A=\left(\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(A=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{2x}{\left(x+2\right)^2}.\frac{\left(x-2\right)\left(x+2\right)}{-x}\)
\(A=\frac{4-2x}{x+2}\)
BT
9 tháng 11 2017
\(A=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right).\left(x^4+\frac{1-x^4}{1+x^2}\right)\)
\(=\left(\frac{\left(x^2-1\right)\left(x^2+1\right)-\left(x^4-x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\right).\left(x^4+\frac{\left(1+x^2\right)\left(1-x^2\right)}{1+x^2}\right)\)
\(=\frac{x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\left(x^4+1-x^2\right)\)
\(=\frac{x^2-2}{x^2+1}\).
30 tháng 1 2022
\(A=\left(\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\cdot\dfrac{2-x}{x}\)
\(=\dfrac{x+2+2x+x-2}{-\left(2-x\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}\)
\(=\dfrac{4x}{-\left(x+2\right)\cdot x}=\dfrac{-4}{x+2}\)
\(A=\frac{2\left|x-4\right|}{x^2+x-20}\)
\(x^2+x-20=x^2-4x+5x-20\)
\(=x\left(x-4\right)+5\left(x-4\right)\)
\(=\left(x-4\right)\left(x+5\right)\)
Trường hợp 1 : \(x-4\ge0\Leftrightarrow x\ge4\)
\(\Rightarrow\)\(A=\frac{2\left(x-4\right)}{\left(x-4\right)\left(x+5\right)}=\frac{2}{x+5}\)
Trường hợp 2 : \(x-4< 0\Leftrightarrow x< 4\)
\(\Rightarrow\)\(A=\frac{2\left(4-x\right)}{\left(x-4\right)\left(x+5\right)}=\frac{-2}{x+5}\)
A = 2|x-4|/(x-4).(x+5)
Nếu x<4 thì |x-4| = -(x+4) => A = -2/x+%
Nếu x>=4 thì |x-4|=x-4 => A = 2/x+5
Vậy ........
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