Có: (3x−5)100+(2x+1)200=((3x−5)50)2+((2x+1)100)2(3x−5)100+(2x+1)200=((3x−5)50)2+((2x+1)100)2 \geq 00
\Rightarrow BPT có nghiệm \Leftrightarrow {3x−5=02y+1=0{3x−5=02y+1=0 \Rightarrow {x=53y=−12{x=53y=−12
Vì \(\hept{\begin{cases}\left(3x-5\right)^{100}\ge0\\\left(2y+1\right)^{200}\ge0\end{cases}\Rightarrow\left(3x-5\right)^{100}+\left(2y+1\right)^{200}\ge0}\)
Theo đề bài:\(\left(3x-5\right)^{100}+\left(2y+1\right)^{200}\le0\)
=>\(\left(3x-5\right)^{100}+\left(2y+1\right)^{200}=0\)
=>\(\hept{\begin{cases}\left(3x-5\right)^{100}=0\\\left(2y+1\right)^{200}=0\end{cases}}\)
=>\(\hept{\begin{cases}3x-5=0\\2y+1=0\end{cases}}\)
=>\(\hept{\begin{cases}3x=5\\2y=-1\end{cases}}\)
=>\(\hept{\begin{cases}x=\frac{5}{3}\\y=\frac{-1}{2}\end{cases}}\)
Vậy \(x=\frac{5}{3}\) và \(y=\frac{-1}{2}\)