cho x,y,z thỏa mãn
2x2+y2+13z2-6xz -4yz-6x+9=0
giá trị của \(\dfrac{2xy+xz-x^2-2y^2-yz}{x^2-y^2}\)
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19 tháng 3 2017
\(2x^2+y^2+13z^2-4yz-6x+9=0\)
\(\Leftrightarrow\left(2x^2-6x+\dfrac{9}{2}\right)+\left(y^2-4yz+4z^2\right)+9z^2+\dfrac{9}{2}=0\)
\(\Leftrightarrow2\left(x^2-3x-\dfrac{9}{4}\right)+\left(y-2z\right)^2+9z^2+\dfrac{9}{2}=0\)
\(\Leftrightarrow2\left(x-\dfrac{3}{2}\right)^2+\left(y-2z\right)^2+9z^2+\dfrac{9}{2}=0\)
Dễ thấy: \(2\left(x-\dfrac{3}{2}\right)^2+\left(y-2z\right)^2+9z^2\ge0\forall x,y,z\)
\(\Rightarrow2\left(x-\dfrac{3}{2}\right)^2+\left(y-2z\right)^2+9z^2+\dfrac{9}{2}\ge\dfrac{9}{2}\forall x,y,z\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}2\left(x-\dfrac{3}{2}\right)^2=0\\\left(y-2z\right)^2=0\\9z^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{3}{2}=0\\y=2z\\z=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=0\\z=0\end{matrix}\right.\)
Khi đó \(P=\dfrac{2\cdot\dfrac{3}{2}\cdot0+\dfrac{3}{2}\cdot0-\left(\dfrac{3}{2}\right)^2-2\cdot0^2-0\cdot0}{\left(\dfrac{3}{2}\right)^2-0^2}=-1\)
19 tháng 3 2017
Đệch, theo đề bài của bn thì Thắng làm đúng òi
Hình như đề thiếu -6xz mới ra -4/5
6 tháng 1 2023
b: 5x^2+5y^2+8xy-2x+2y+2=0
=>4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0
=>(x-1)^2+(y+1)^2+(2x+2y)^2=0
=>x=1 và y=-1
M=(1-1)^2015+(1-2)^2016+(-1+1)^2017=1
9 tháng 9 2021
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow\dfrac{xy+yz+xz}{xyz}=0\Leftrightarrow xy+yz+xz=0\Leftrightarrow yz=-xy-xz\)
Ta có \(x^2+2yz=x^2+yz-xy-xz=\left(x-y\right)\left(x-z\right)\)
Tương tự \(y^2+2xz=\left(y-x\right)\left(y-z\right);z^2-2xy=\left(z-x\right)\left(z-y\right)\)
\(A=\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(y-z\right)\left(y-x\right)}+\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\\ A=\dfrac{-yz\left(y-z\right)-xz\left(z-x\right)-xy\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{-yz\left(y-z\right)+xz\left(y-z\right)+xz\left(x-y\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{\left(y-z\right)\left(xz-yz\right)+\left(x-y\right)\left(xz-xy\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 3 2021
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
f(x,y,z) =\(\left(x^2+9z^2-6xz\right)+\left(y^2+4z^2-4yz\right)+\left(x^2-6x+9\right)\)
\(f\left(x,y,z\right)=\left(x-3z\right)^2+\left(y-2z\right)^2+\left(x-3\right)^2\)
\(f\left(x,y,z\right)\ge0\forall x,y,z\in R\)
\(f\left(x,y,z\right)=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\x-3z=0\\y-2z=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=3z\\y=2z\end{matrix}\right.\\xy=6z^2\\x^2=9z^2\\y^2=4z^2\end{matrix}\right.\)
\(A=\dfrac{2xy+xz-x^2-2y^2-yz}{x^2-y^2}=\dfrac{12z^2+3z^2-9z^2-8z^2-2z^2}{9z^2-4z^2}=\dfrac{-4z^2}{5z^2}=-\dfrac{4}{5}\)