+ Áp dụng bđt \(a^2+b^2+c^2\ge ab+bc+ca\) ta có
\(2x^4+2y^4+2z^4=2\left(x^4+y^4+z^4\right)\ge2\left(x^2y^2+y^2z^2+z^2x^2\right)\ge2\left(xy^2z+x^2yz+xyz^2\right)\)
\(=2xyz\left(x+y+z\right)=2xyz\)
Dấu "=" \(\Leftrightarrow x=y=z=\frac{1}{3}\)
+ Áp dụng bđt \(a^2+b^2+c^2\ge ab+bc+ca\) ta có
\(2x^4+2y^4+2z^4=2\left(x^4+y^4+z^4\right)\ge2\left(x^2y^2+y^2z^2+z^2x^2\right)\ge2\left(xy^2z+x^2yz+xyz^2\right)\)
\(=2xyz\left(x+y+z\right)=2xyz\)
Dấu "=" \(\Leftrightarrow x=y=z=\frac{1}{3}\)
Giaỉ hệ phương trình \(\left\{{}\begin{matrix}x+y+z=1\\2x+2y-2xy+z^2=1\end{matrix}\right.\)
giải hệ phương trình
a)\(\left\{{}\begin{matrix}\left(x^2+1\right)\left(y^2+1\right)=10\\\left(x+y\right)\left(xy-1\right)=3\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}x^2+y^2+2\left(xy-2\right)=0\\x^2+y^2-2xy=16\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2-2x\sqrt{y}+2y=x\\y^2-2y\sqrt{x}+2z=y\\z^2-2z\sqrt{x}+2x=z\end{matrix}\right.\)
Giải hệ phương trình sau: \(\left\{{}\begin{matrix}\sqrt{x}\left(1+y\right)=2y\\\sqrt{y}\left(1+z\right)=2z\\\sqrt{z}\left(1+x\right)=2x\end{matrix}\right.\)
Giải hệ phương trình :
1, \(\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{4}{x}+\frac{1}{y-2}=1\end{matrix}\right.\)
2 , \(\left\{{}\begin{matrix}\frac{2}{2x-y}-\frac{1}{x+y}=0\\\frac{3}{2x-y}-\frac{6}{x+y}=-1\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-2y\right)-15\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}2x+y=7\\-x+4y=10\end{matrix}\right.\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}2x^2=y\left(x^2+1\right)\\2y^2=z\left(y^2+1\right)\\2z^2=x\left(z^2+1\right)\end{matrix}\right.\)
giải hệ phương trình
a) \(\left\{{}\begin{matrix}3x-2y+z=14\\2x+y-z=3\\z-2x=-5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2-y^2=4y+2x+3\\x^2+2x+y=0\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\left|xy-4\right|=8-y^2\\xy=2+x^2\end{matrix}\right.\)
Giải hệ phương trình:
a, \(\left\{{}\begin{matrix}\sqrt{2x+3}+\sqrt{4-y}=4\\\sqrt{2y+3}+\sqrt{4-y}=4\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}2y=\dfrac{y^2+1}{x^2}\\2x=\dfrac{x^2+1}{y^2}\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}2x^2+y^2-3xy=12\\2\left(x+y\right)^2-y^2=14\end{matrix}\right.\)
Giải hệ phương trình : \(\left\{{}\begin{matrix}x\left(yz+1\right)=2z\\y\left(zx+1\right)=2x\\z\left(xy+1\right)=2y\end{matrix}\right.\)
giải hệ phương trình sau
\(\left\{{}\begin{matrix}2x-3y=4\\x+2y=-5\end{matrix}\right.\)