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6 tháng 9 2021
a) x2+y2-4x+4y+8=0
⇔ (x-2)2+(y+2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
b)5x2-4xy+y2=0
⇔ x2+(2x-y)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
c)x2+2y2+z2-2xy-2y-4z+5=0
⇔ (x-y)2+(y-1)2+(z-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)
6 tháng 9 2021
b: Ta có: \(5x^2-4xy+y^2=0\)
\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)
\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
LT
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21 tháng 1 2022
b: \(=\dfrac{12\left(y-z\right)^4+3\left(y-z\right)^5}{6\left(y-z\right)^2}=2\left(y-z\right)^2+\dfrac{1}{2}\left(y-z\right)^3\)
29 tháng 6 2019
a) \(\Leftrightarrow4x^2+2y^2+4xy-20x-8y+26=0\)
\(\Leftrightarrow4x^2+4x\left(y-5\right)+\left(y-5\right)^2-\left(y-5\right)^2+2y^2-8y+26=0\)
\(\Leftrightarrow\left(2x+y-5\right)^2+y^2+2y+1=0\)
\(\Leftrightarrow\left(2x+y-5\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+y-5=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\) ( TM )
b) \(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+6y+9\right)+\left(z^2-2z+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2+\left(y+3\right)^2+\left(z-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+3=0\\z-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-3\\z=1\end{matrix}\right.\) ( TM )
c) \(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2xz\right)+\left(x^2+2x+1\right)+\left(z^2-4z+4\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+1\right)^2+\left(z-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=0\\x+1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-1\\z=2\end{matrix}\right.\) ( TM )
\(x^2-2x+y^2+4y+5+\left(2z-3\right)^2=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+\left(2z-3\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2+\left(2z-3\right)^2=0\)
Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left(y+2\right)^2\ge0\\\left(2z-3\right)^2\ge0\end{cases}}\) nên \(\left(x-1\right)^2+\left(y+2\right)^2+\left(2z-3\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\\\left(2z-3\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=2\\z=\frac{3}{2}\end{cases}}}\)