Giải hệ phương trình sau: \(\left\{\begin{array}{l} \frac{x}{2}-\frac{y}{3}=0 \\ \frac{4}{y+4}=\frac{9}{x+8} \end{array}\right.\).

* Hướng dẫn giải

ĐK: \(x \neq-8 ; y \neq-4\)

Ta có:

\(\left\{\begin{array}{l} \frac{x}{2}-\frac{y}{3}=0 \\ \frac{4}{y+4}=\frac{9}{x+8} \end{array} \Leftrightarrow\left\{\begin{array}{l} 3 x-2 y=0 \\ 4(x+8)=9(y+4) \end{array}\right.\right.\)

\(\Leftrightarrow\left\{\begin{array}{l} 3 x-2 y=0 \\ 4(x+8)=9(y+4) \end{array} \Leftrightarrow\left\{\begin{array}{l} 3 x-2 y=0 \\ 4 x+32=9 y+36 \end{array} \Leftrightarrow\left\{\begin{array}{l} x=\frac{2}{3} y \\ 4 x-9 y=4 \end{array}\right.\right.\right.\)

\(\Leftrightarrow\left\{\begin{array}{l} x=\frac{2}{3} y \\ 4 x-9 y=4 \end{array} \Leftrightarrow\left\{\begin{array}{l} x=\frac{2}{3} y \\ 4 \cdot\left(\frac{2}{3} y\right)-9 y=4 \end{array} \Leftrightarrow\left\{\begin{array}{l} x=-\frac{8}{19} \\ y=-\frac{12}{19} \end{array}\right.\right.\right.\)

Vậy nghiệm của hệ phương trình là: \(\left(-\frac{8}{19} ;-\frac{12}{19}\right)\)