Question

giải hpt

\(\left\{{}\begin{matrix}6x+6y=5xy\\\dfrac{4}{x}-\dfrac{3}{y}=1\end{matrix}\right.\)

Answer 1

Lời giải:
Ta có:

HPT \(\Leftrightarrow \left\{\begin{matrix} 6x+6y=5xy\\ 4y-3x=xy\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} 6x+6y=5xy\\ 20y-15x=5xy\end{matrix}\right.\)

Lấy PT(1) - PT(2):

\(6x+6y-(20y-15x)=0\)

\(\Leftrightarrow 21x=14y\Leftrightarrow 3x=2y\Rightarrow y=1,5x\)

Thay vào PT ban đầu:

\(6x+6.1,5x=5x.1,5x\)

\(\Leftrightarrow 15x=7,5x^2\Leftrightarrow x(7,5x-15)=0\)

Vì $x\neq 0$ nên \(7,5x-15=0\Leftrightarrow x=2\Rightarrow y=1,5.2=3\)

Vậy $(x,y)=(2,3)$