Ta có
\(\left(\dfrac{1}{3}-2x\right)^{102}\ge0\forall x\\ \left(3y-x\right)^{104}\ge0\forall x\)
Để \(\left(\dfrac{1}{3}-2x\right)^{102}+\left(3y-x\right)^{104}=0\)
\(\left\{{}\begin{matrix}\left(\dfrac{1}{3}-2x\right)^{102}=0\\\left(3y-x\right)^{104}=0\end{matrix}\right.=>\left\{{}\begin{matrix}\dfrac{1}{3}-2x=0\\3y-x=0\end{matrix}\right.=>\left\{{}\begin{matrix}x=\dfrac{1}{6}\\3y-\dfrac{1}{6}=0\end{matrix}\right.=>\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=\dfrac{1}{18}\end{matrix}\right.\)
`(1/3-2x)^102+(3y-x)^104=0`
Vì `(1/3-2x)^102 >= 0 AA x`
`(3y-x)^104 >= 0 AA x,y`
`=>{((1/3-2x)^102=0),((3y-x)^104=0):}`
`<=>{(1/3-2x=0),(3y-x=0):}`
`<=>{(x=1/6),(3y-1/6=0):}<=>{(x=1/6),(y=/18):}`
