a, \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{3}=\frac{5y}{10}\)
Áp dụng tính chất Dãy tỉ số bằng nhau
\(\frac{x}{3}=\frac{5y}{10}=\frac{x-5y}{3-10}=\frac{2,1}{-7}\frac{-3}{10}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{3}=-\frac{3}{10}\\\frac{y}{2}=-\frac{3}{10}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{9}{10}\\y=-\frac{3}{5}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-\frac{9}{10}\\y=-\frac{3}{5}\end{matrix}\right.\)
b, Đặt \(\frac{x}{3}=\frac{y}{7}=\frac{z}{2}=k\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{3}=k\\\frac{y}{7}=k\\\frac{z}{2}=k\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3k\\y=7k\\z=2k\end{matrix}\right.\)
2x2+y2+3z2= 2.(3k)2+(7k)2+3.(2k)2
316= 18k2+49k2+12k2
316=k2.(18+49+12)
316=79k2
4=k2
\(\Rightarrow\left[{}\begin{matrix}k=-2\\k=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3k\\y=7k\\z=2k\end{matrix}\right.\\\left\{{}\begin{matrix}x=3k\\y=7k\\z=2k\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-6\\y=-14\\z=-4\end{matrix}\right.\\\left\{{}\begin{matrix}x=6\\y=14\\z=4\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-6\\y=-14\\z=-4\end{matrix}\right.\)
và \(\left\{{}\begin{matrix}x=6\\y=14\\z=4\end{matrix}\right.\)
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Bài giải
a, \(2x=3y\text{ }\Rightarrow\text{ }\frac{x}{3}=\frac{y}{2}=\frac{5y}{10}=\frac{x-5y}{3-10}=\frac{2,1}{-7}=\frac{-3}{10}\)
( Áp dụng t/c ... )
\(\Rightarrow\text{ }x=\frac{-3}{10}\cdot3=-\frac{9}{10}\)
\(y=\frac{-3}{10}\cdot2=-\frac{6}{10}\)
b, \(\frac{x}{3}=\frac{y}{7}=\frac{z}{2}\Rightarrow\text{ }\frac{x^2}{9}=\frac{y^2}{49}=\frac{z^2}{4}\text{ }=\frac{2x^2}{18}=\frac{3z^2}{12}=\frac{2x^2+y^2+3z^2}{18+49+12}=\frac{316}{79}=4\)
\(\Rightarrow\text{ }x^2=4\cdot9=36\text{ }\Rightarrow\text{ }x=\pm6\)
\(y^2=4\cdot49=196\text{ }\Rightarrow\text{ }y=\pm14\)
\(z^2==4\cdot4=16\text{ }\Rightarrow\text{ }z=4\)
