Giả sử \(x\ge y\ge z\ge t\)
Có 5(x+y+z+t) = 2xyzt
<=> \(2=\dfrac{5}{yzt}+\dfrac{5}{xyz}+\dfrac{5}{xyt}+\dfrac{5}{xzt}+\dfrac{10}{xyzt}\le\dfrac{20}{t^3}+\dfrac{10}{t^4}\le\dfrac{30}{t^3}\)
<=> t3 \(\le15\)
<=> \(\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)
TH1: t = 1
<=> \(2=\dfrac{5}{yz}+\dfrac{5}{xyz}+\dfrac{5}{xy}+\dfrac{5}{xz}+\dfrac{10}{xyz}=\dfrac{5}{xy}+\dfrac{5}{yz}+\dfrac{5}{zx}+\dfrac{15}{xyz}\)
\(\le\dfrac{15}{z^2}+\dfrac{15}{z^3}\le\dfrac{30}{z^2}\)
<=> z2 \(\le15\)
<=> \(\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)
- Với z = 1
PT <=> 5 (x+y+2) + 10 = 2xy
<=> (2x-5)(2y-5) = 65
<=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=35\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=9\\y=5\end{matrix}\right.\end{matrix}\right.\)
Vậy (x;y;z;t) = (35;3;1;1) hoặc (9;5;1;1) và có hoán vị
- Với z = 2;3 => Vô nghiệm
TH2: t = 2
PT <=> 5(x+y+z) + 20 = 4xyz
<=> \(4=\dfrac{5}{xy}+\dfrac{5}{yz}+\dfrac{5}{zx}+\dfrac{20}{xyz}\le\dfrac{35}{z^2}\)
<=> \(\left[{}\begin{matrix}z=1\left(l\right)\\z=2\left(c\right)\end{matrix}\right.\)
<=> 5(x+y+4) + 10 = 8xy
<=> (8x-5)(8y-5) = 265
=> Vô nghiệm
KL: Vậy (x;y;z;t) = (35;3;1;1) hoặc (9;5;1;1) và có hoán vị
