Question

\(\dfrac{x}{5}=\dfrac{4}{-3}\) và x\(^2\)+y\(^2\)=34

\(\dfrac{x}{4}=\dfrac{y}{7}\) và x\(^2\)-\(\)y\(^2\)= -33

 

Answer 1

\(\dfrac{x}{5}=\dfrac{4}{-3}\)

⇔\(-3.x=4.5\)

⇔\(-3x=20\)

⇔\(x=-\dfrac{20}{3}\)

Answer 2

b: Đặt \(\dfrac{x}{4}=\dfrac{y}{7}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=7k\end{matrix}\right.\)

Ta có: \(x^2-y^2=-33\)

\(\Leftrightarrow k^2=1\)

Trường hợp 1: k=1

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=4\\y=7k=7\end{matrix}\right.\)

Trường hợp 2: k=-1

\(\Leftrightarrow\left\{{}\begin{matrix}x=4k=-4\\y=7k=-7\end{matrix}\right.\)