Question

(3x+2)(2y-3)=6xy

(4x+5)(y-5)=4xy

Giải hệ pt ak, em ko đánh dấu ngoặc đc

Answer 1

\(\left\{{}\begin{matrix}\left(3x+2\right)\left(2y-3\right)=6xy\\\left(4x+5\right)\left(y-5\right)=4xy\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\text{6xy+4y-9x-6=6xy}\\\text{4xy+5y-20x-25=4xy}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\text{4y-9x=6}\\\text{5y-20x=25}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\text{20y-45x=30}\\\text{20y-80x=100}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\text{35x=-70}\\\text{4y-9x=6}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\text{x=-2}\\\text{4y-9.(-2)=6}\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}\text{x=-2}\\\text{y=-3}\end{matrix}\right.\)

vậy ...

 

Answer 2

Ta có: \(\left\{{}\begin{matrix}\left(3x+2\right)\left(2y-3\right)=6xy\\\left(4x+5\right)\left(y-5\right)=4xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6xy-9x+4y-6-6xy=0\\4xy-20x+5y-25-4xy=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-9x+4y=6\\-20x+5y=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-180x+80y=120\\-180x+45y=225\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}35y=-105\\-9x+4y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\-9x=6-4y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-9x=6-4\cdot\left(-3\right)=6+12=18\\y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là (x,y)=(-2;-3)