Giải phương trình \(5x^2+4y^2+2=4xy+2x+4y\\\)
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Ta có K = (x2 + 4y2 + 1 - 4xy - 2x + 4y) + (4x2 + 4x + 1) + 1 = (2y - x + 1)2 + (2x + 1)2 + 1 >= 1
Vậy GTNN là -1 đạt được tại x = -0,5; y = - 0,25
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\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y-x\right)\left(2y+x\right)}{\left(x-2y\right)^2}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)
Điều kiện: \(x\ne2y;x\ne-2y;x\ne0;y\ne0\)
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y+x\right)}{\left(x-2y\right)}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\times\frac{x-2y}{x+2y}\times\frac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}=\frac{2\left(x-2y\right)}{5y}\)
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\(=\dfrac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\cdot\dfrac{\left(x-2y\right)^2}{-\left(x-2y\right)\left(x+2y\right)}:\dfrac{5x^2y-10xy^2}{x^3+6x^2y+12xy^3+8y^3}\)
\(=\dfrac{-2x\left(x-2y\right)^2}{\left(x+2y\right)^3}\cdot\dfrac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}\)
\(=\dfrac{-2x\cdot\left(x-2y\right)}{5xy}=\dfrac{-2\left(x-2y\right)}{5y}\)
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mk sửa lại đoạn sau:
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x-1=0\\2z-x-1=0\\2y+x-z-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\2z-2=0\\2y-z=0\left(x-1=0\right)\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\z=1\\2y=1\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\z=1\\y=\dfrac{1}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)