\(3x^2+y^2+z^2+2x-2y+2xy+3=0\),x=?,y=?,z=?
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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 \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
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10 tháng 12 2021
\(a,=xy\left(x+2y+1\right)\\ b,=x^2\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x-2\right)\left(x+2\right)\\ c,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ d,=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2=\left(x-2\right)\left(x+2+x-2\right)=2x\left(x-2\right)\\ e,=\left(x+1\right)^2-y^2=\left(x+y+1\right)\left(x-y+1\right)\\ g,=\left(x+9-6x\right)\left(x+9+6x\right)=\left(9-5x\right)\left(7x+9\right)\\ h,=\left(x-y\right)^2-\left(z-t\right)^2=\left(x-y-z+t\right)\left(x-y+z-t\right)\\ i,=\left(x-1\right)^3-y^3=\left(x-y-1\right)\left(x^2-2x+1+xy+y+y^2\right)\)
19 tháng 9 2019
\(a,4x^2+9y^2+4x-24y+17=0\)
\(\Rightarrow\left(4x^2+4x+1\right)+\left(9y^2-24y+16\right)=0\)
\(\Rightarrow\left(2x+1\right)^2+\left(3y-4\right)^2=0\)
\(\left(2x+1\right)^2\ge0;\left(3y-4\right)^2\ge0\)
\(\Rightarrow\hept{\begin{cases}\left(2x+1\right)^2=0\\\left(3y-4\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+1=0\\3y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{4}{3}\end{cases}}}\)
Lời giải:
\(3x^2+y^2+z^2+2x-2y+2xy+3=0\)
\(\Leftrightarrow (x^2+y^2+1+2xy-2y-2x)+2(x^2+2x+1)+z^2=0\)
\(\Leftrightarrow (x+y-1)^2+2(x+1)^2+z^2=0\)
Vì \(\left\{\begin{matrix} (x+y-1)^2\geq 0\\ (x+1)^2\geq 0\\ z^2\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{R}\)
Do đó: \((x+y-1)^2+2(x+1)^2+z^2\geq 0\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} (x+y-1)^2=0\\ (x+1)^2=0\\ z^2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=-1\\ y=2\\ z=0\end{matrix}\right.\)