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\(\left\{{}\begin{matrix}x^4+y^4\ge2x^2y^2\\y^4+z^4\ge2y^2z^2\\x^4+z^4\ge2x^2z^2\end{matrix}\right.\) \(\Rightarrow x^4+y^4+z^4\ge x^2y^2+y^2z^2+x^2z^2\)
Lại có:
\(\left\{{}\begin{matrix}x^2y^2+y^2z^2\ge2xy^2z\\x^2y^2+x^2z^2\ge2x^2yz\\y^2z^2+x^2z^2\ge2xyz^2\end{matrix}\right.\) \(\Rightarrow x^2y^2+y^2z^2+x^2z^2\ge xy^2z+x^2yz+xyz^2\)
\(\Rightarrow x^2y^2+y^2z^2+x^2z^2\ge xyz\left(x+y+z\right)=xyz\)
\(\Rightarrow x^4+y^4+z^4\ge xyz\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)
\(\Rightarrow\) Hệ có nghiệm duy nhất \(\left(x;y;z\right)=\left(\dfrac{1}{3};\dfrac{1}{3};\dfrac{1}{3}\right)\)
b, Ta có : \(\left\{{}\begin{matrix}x+y+z=3\\y+z+t=4\\z+t+x=5\\t+x+y=6\end{matrix}\right.\)
=> \(x+y+z+y+z+t+z+t+x+t+x+y=18\)
=> \(3\left(x+y+z+t\right)=18\)
=> \(x+y+z+t=6\)
=> \(x+y+z+t=x+y+t\)
=> \(z=0\)
=> \(\left\{{}\begin{matrix}x+y=3\\y+t=4\\x+t=5\\x+y+t=6\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x+y=3\\y+t=4\\x+t=5\\y+5=6\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x+1=3\\t+1=4\\x+t=5\\y=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2\\t=3\\x+t=5\\y=1\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=2\\y=1\\z=0\\t=3\end{matrix}\right.\)
a, Ta có : \(\left\{{}\begin{matrix}7xy=12\left(x+y\right)\\9yz=20\left(y+z\right)\\8zx=15\left(z+x\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}7xy-12x-12y=0\\9yz-20y-20z=0\\8zx-15z-15x=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{12y}{7y-12}\\y=\frac{20z}{9z-20}\\x=\frac{15z}{8z-15}\end{matrix}\right.\)
=> \(12y\left(8z-15\right)=15z\left(7y-12\right)\)
=> \(96yz-180y=105yz-180z\)
=> \(105yz-96yz=-180y+180z\)
=> \(9yz=-180y+180z\)
=> \(180z-180y=20y+20z\)
=> \(180z-20z=180y+20y=160z=200y\)
=> \(y=\frac{4}{5}z\)
=> \(\frac{20z}{9z-20}=\frac{4z}{5}\)
=> \(4z\left(9z-20\right)=100z\)
=> \(36z^2-180z=0\)
=> \(\left[{}\begin{matrix}z=5\\z=0\end{matrix}\right.\)
TH1 : z = 0 .
=> \(x=y=z=0\)
TH2 : z = 5 .
=> \(\left\{{}\begin{matrix}7xy=12\left(x+y\right)\\45y=20\left(y+5\right)\\40x=15\left(5+x\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=3\\y=4\\z=5\end{matrix}\right.\)
\(x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyz\left(x+y+z\right)=xyz\)
Dấu = xảy ra khi \(x=y=z=\dfrac{1}{3}\)