Question

[Bùi Anh Nhật]

 giúp e vớiii

Answer 1

[Nguyễn Lê Phước Thịnh]

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{4}{5y}=7\\\dfrac{3}{4x}-\dfrac{2}{5y}=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{4}{5y}=7\\\dfrac{6}{4x}-\dfrac{4}{5y}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{6}{4x}=7+6\\\dfrac{2}{3x}+\dfrac{4}{5y}=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{3}{2x}=13\\\dfrac{2}{3x}+\dfrac{4}{5y}=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x}\left(\dfrac{2}{3}+\dfrac{3}{2}\right)=13\\\dfrac{2}{3x}+\dfrac{4}{5y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}\cdot\dfrac{13}{6}=13\\\dfrac{2}{3x}+\dfrac{4}{5y}=7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x}=13:\dfrac{13}{6}=6\\\dfrac{2}{3x}+\dfrac{4}{5y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\\dfrac{4}{5y}=7-\dfrac{2}{3x}=7-\dfrac{2}{3\cdot\dfrac{1}{6}}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{6}\\\dfrac{4}{5y}=7-\dfrac{2}{\dfrac{1}{2}}=7-2\cdot2=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{6}\\5y=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=\dfrac{4}{15}\end{matrix}\right.\left(nhận\right)\)