Giải hệ phương trình:
a, \(\left\{{}\begin{matrix}\frac{x}{2}=\frac{y}{3}\\\frac{x+8}{y+4}=\frac{9}{4}\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}0,75x-3,2y=10\\x\sqrt{3}-y\sqrt{2}=4\sqrt{3}\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}\frac{2x+3}{y-1}=\frac{4x+1}{2y+1}\\\frac{x+2}{y-1}=\frac{x-4}{y+2}\end{matrix}\right.\)
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a) \(\left\{{}\begin{matrix}\frac{x}{2}=\frac{y}{3}\\\frac{x+8}{y+4}=\frac{9}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{2y}{3}\\\frac{\frac{2y}{3}+8}{y+4}=\frac{9}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{-12}{19}\\x=\frac{-8}{19}\end{matrix}\right.\)
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b) \(\left\{{}\begin{matrix}0,75x-3,2y=10\\x\sqrt{3}-y\sqrt{2}=4\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3,2y+10}{0,75}\\\frac{\left(3,2y+10\right)\sqrt{3}}{0,75}-y\sqrt{2}=4\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{\frac{16\sqrt{3}}{5}y+10\sqrt{3}-\frac{3\sqrt{2}}{4}y}{0,75}=4\sqrt{3}\\x=\frac{3,2y+10}{0,75}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y\left(\frac{16\sqrt{3}}{5}-\frac{3\sqrt{2}}{4}\right)+10\sqrt{3}=3\sqrt{3}\\x=\frac{3,2y+10}{0,75}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{-140\sqrt{3}}{64\sqrt{3}-15\sqrt{2}}\\x=\frac{\frac{-448\sqrt{3}}{64\sqrt{3}-15\sqrt{2}}+10}{0,75}\end{matrix}\right.\)
Nghiệm đẹp lắm.
c) \(\left\{{}\begin{matrix}\frac{2x+3}{y-1}=\frac{4x+1}{2y+1}\\\frac{x+2}{y-1}=\frac{x-4}{y+2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+3\right)\left(2y+1\right)-\left(y-1\right)\left(4x+1\right)=0\\\left(x+2\right)\left(y+2\right)-\left(y-1\right)\left(x-4\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x+5y+4=0\\3x+6y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2y\\-12y+5y+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{4}{7}\\x=\frac{-8}{7}\end{matrix}\right.\)
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