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\(\frac{4xy+2\left(x+y\right)\left(x-y\right)}{2\left(x+y\right)\left(x-y\right)}\)
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14 tháng 1 2018
Do \(x+y+z=0\)
\(\Rightarrow x=-\left(y+z\right)\Rightarrow x^2=\left(y+z\right)^2\Rightarrow4yz-x^2=4yz-\left(y+z^2\right)=-\left(y-z\right)^2\)
Tương tự \(4zx-y^2=-\left(z-x\right)^2\)
\(4xy-z^2=-\left(x-y\right)^2\)
Ta lại có: \(yz+2x^2=yz+x^2-x\left(y+z\right)=yz+x^2-xy-xz=\left(x-y\right)\left(x-z\right)\)
Tương tự: \(zx+2y^2=\left(y-x\right)\left(y-z\right)\)
\(xy+2z^2=\left(y-z\right)\left(y-y\right)\)
\(P=\frac{\left(4yz-x^2\right)\left(4zx-y^2\right)\left(4xy-z^2\right)}{\left(yz+2x^2\right)\left(zx+2y^2\right)\left(xy+2z^2\right)}=\frac{-\left(y-z\right)^2\left(z-x\right)^2\left(x-y^2\right)}{\left(x-y\right)\left(x-z\right)\left(y-x\right)\left(y-z\right)\left(z-x\right)\left(z-y\right)}\)
\(=\frac{-\left(y-z\right)^2\left(z-x\right)^2\left(x-y\right)^2}{-\left(y-z\right)^2\left(z-x\right)^2\left(x-y\right)^2}=1\)
23 tháng 7 2023
a) \(x^2y\left(5xy-2x^2y-y^2\right)\)
\(=5x^3y^2-2x^4y^2-x^2y^3\)
b) \(\left(x-2y\right)\left(2x^3+4xy\right)\)
\(=2x^4+4x^2y-4x^3y-8xy^2\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 9 2020
ĐKXĐ: ...
\(\left\{{}\begin{matrix}x+y+\frac{1}{x+y}+x-y=1\\5\left(x+y\right)^2+3\left(x-y\right)^2+\frac{5}{\left(x+y\right)^2}=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+\frac{1}{x+y}+x-y=1\\5\left(x+y+\frac{1}{x+y}\right)^2+3\left(x-y\right)^2=23\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y+\frac{1}{x+y}=a\\x-y=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=1\\5a^2+3b^2=23\end{matrix}\right.\)
\(\Rightarrow5a^2+3\left(1-a\right)^2-23=0\)
\(\Leftrightarrow...\)
\(\frac{4xy+2\left(x+y\right)\left(x-y\right)}{2\left(x+y\right)\left(x-y\right)}\)
\(=\frac{4xy+2\left(x^2-y^2\right)}{2\left(x^2-y^2\right)}\)
\(=\frac{2\left(x^2+2xy-y^2\right)}{2\left(x^2-y^2\right)}=\frac{x^2+2xy-y^2}{x^2-y^2}\)