Question

[Ryan Nguyễn]

 Tìm x và y biết rằng:
(3x−5)¹⁰⁰+ (2y+1)²⁰⁰ < hoặc = 0

Answer 1

[Ryan Nguyễn]

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

Answer 2

[Trà My]

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 \(y=\frac{-1}{2}\)