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giải hệ phương trình 1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\) 2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\) 3 ,...
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giải hệ phương trình

1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)

2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)

3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)

4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)

5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)

6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)

7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)

8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)

9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)

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11 tháng 12 2019

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8 tháng 8 2018

\(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)

\(=\frac{3x+2}{9x^2-4}-\frac{3x-2}{9x^2-4}+\frac{3x-6}{9x^2-4}\)

\(=\frac{3x+2-3x+2+3x-6}{9x^2-4}\)

\(=\frac{3x-2}{9x^2-4}\)

\(=\frac{1}{3x+2}\)

\(\frac{18}{\left(x-3\right)\left(x^2-9\right)}-\frac{3}{x^2-6x+9}-\frac{x^2}{x^2-9}\)

\(=\frac{18}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\) \(-\frac{3\left(x+3\right)}{\left(x-3\right)\left(x-3\right)\left(x+3\right)}\)\(-\frac{x^2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x-3\right)}\)

\(=\frac{18-3x-9-x^3+3x^2}{\left(x-3\right)^2\left(x+3\right)}\)

\(=\frac{-x^3+3x^2-3x+9}{\left(x-3^2\right)\left(x+3\right)}\)

\(=\frac{\left(-x^2-3\right)\left(x-3\right)}{\left(x-3^2\right)\left(x+3\right)}\)

\(=\frac{-x^2-3}{\left(x-3\right)\left(x+3\right)}\)

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