Question

x^3/8=y^3/64và x^2+y^2=20

giúp em vs ạ

 

Answer 1

\(\dfrac{x^3}{8}=\dfrac{y^3}{64}\)

\(\Rightarrow\dfrac{x^3}{2^3}=\dfrac{y^3}{4^3}=\left(\dfrac{x}{2}\right)^3=\left(\dfrac{y}{4}\right)^3\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}\)

Đặt: \(\dfrac{x}{2}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=4k\end{matrix}\right.\)

Mà: \(x^2+y^2=20\)

\(\Rightarrow\left(2k\right)^2+\left(4k\right)^2=20\)

\(\Rightarrow4k^2+16k^2=20\)

\(\Rightarrow20k^2=20\)

\(\Rightarrow k^2=1\)

\(\Rightarrow k=\pm1\)

Khi \(k=1\Rightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

khi \(k=-1\Rightarrow\left\{{}\begin{matrix}x=-2\\y=-4\end{matrix}\right.\)