1+1+1

=3

bóng chày

nhảy cao

24 100%

\(\left(\frac{1}{7}x-\frac{1}{3}\right).\left(-\frac{1}{5}x+\frac{3}{5}\right)=0\)

=>\(\frac{1}{7}x-\frac{1}{3}=0=>\frac{1}{7}x=\frac{1}{3}=>x=\frac{7}{3}\)

\(-\frac{1}{5}x+\frac{3}{5}=0=>-\frac{1}{5}x=-\frac{3}{5}\) =>\(x=3\)

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 \(y=\frac{-1}{2}\)