Question

Giải hệ phương trình :

\(\left\{{}\begin{matrix}4x+\dfrac{9y}{4}=120\\\dfrac{9y}{4x}=\dfrac{4x}{y}\end{matrix}\right.\)

Answer 1

ĐK: \(x.y\ne0\)

\(\dfrac{9y}{4x}=\dfrac{4x}{y}\Leftrightarrow9y^2=16x^2\Leftrightarrow\left[{}\begin{matrix}4x=3y\\4x=-3y\end{matrix}\right.\)

TH1: \(4x=3y\) thay vào pt đầu:

\(3y+\dfrac{9y}{4}=120\Leftrightarrow\dfrac{21y}{4}=120\Rightarrow\left\{{}\begin{matrix}y=\dfrac{160}{7}\\x=\dfrac{120}{7}\end{matrix}\right.\)

TH2: \(4x=-3y\) thay vào pt đầu:

\(-3y+\dfrac{9y}{4}=120\Leftrightarrow\dfrac{-3y}{4}=120\Rightarrow\left\{{}\begin{matrix}y=-160\\x=120\end{matrix}\right.\)