a)\(\frac{x}{5}=\frac{y}{-3}\Rightarrow\frac{x^2+y}{5^2.-3}=\frac{34}{-125}\)
\(\Rightarrow\frac{x}{5}=-\frac{34}{125}\Rightarrow x=-\frac{34}{125}.5=-\frac{34}{25}\)
\(\Rightarrow\frac{y}{-3}=-\frac{34}{125}\Rightarrow y=-\frac{34}{125}.-3=\frac{102}{125}\)
b)\(4x=-5y\Rightarrow\frac{4x}{20}=-\frac{5y}{20}\Rightarrow\frac{x}{5}=\frac{y}{-4}=K\)
\(\frac{x}{5}=K\Rightarrow x=5K;\frac{y}{-4}=K\Rightarrow y=-4K\)
\(x.y=-80\)
\(5K.-4K=-80\)
\(K^2.\left(-4.5\right)=-80\)
\(K^2=-80:\left(-20\right)\)
\(K^2=4\Rightarrow K=2\)
\(\frac{x}{5}=2\Rightarrow x=10\)
\(\frac{y}{-4}=2\Rightarrow y=-8\)
a, Đặt \(\hept{\begin{cases}x=5k\\y=-3k\end{cases}}\)Theo bài ra ta có : \(x^2+y=34\)
\(\left(5k\right)^2-3k=34\Leftrightarrow25k^2-3k=34\Leftrightarrow k\left(25k-3\right)=34\)
\(\Leftrightarrow\orbr{\begin{cases}k=34\\25k-3=34\end{cases}\Leftrightarrow\orbr{\begin{cases}k=34\\k=\frac{37}{25}\end{cases}}}\)
b, Theo bài ra ta có : \(4x=-5y\Leftrightarrow\frac{x}{-5}=\frac{y}{4}\)
Đặt \(\hept{\begin{cases}x=-5k\\y=4k\end{cases}}\)Theo bài ra ta có : \(xy=-80\)
\(\Leftrightarrow-5k.4k=-80\Leftrightarrow-20k^2=-80\Leftrightarrow k^2=4\Leftrightarrow k=\pm2\)
Với k = 2 : \(\hept{\begin{cases}x=-10\\y=8\end{cases}}\)Với k = -2 \(\hept{\begin{cases}x=10\\y=-8\end{cases}}\)
