a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x+2y}{7+2\cdot5}=\dfrac{51}{17}=3\)
Do đó: x=21; y=15
a) \(\Rightarrow\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{2y}{10}=\dfrac{x+2y}{7+10}=\dfrac{51}{17}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.7=21\\y=3.5=15\end{matrix}\right.\)
b) \(\dfrac{x}{5}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{16}=\dfrac{x^2-y^2}{25-16}=\dfrac{1}{9}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{25}{9}\\y^2=\dfrac{16}{9}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=\dfrac{4}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\end{matrix}\right.\)
c) \(\dfrac{x}{y}=\dfrac{2}{5}\Rightarrow\dfrac{x}{2}=\dfrac{y}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)
\(\Rightarrow xy=10k^2=40\Rightarrow k=\pm2\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4\\y=10\end{matrix}\right.\\\left\{{}\begin{matrix}x=-4\\y=-10\end{matrix}\right.\end{matrix}\right.\)
\(a,5x=7y\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{5}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x+2y}{7+10}=\dfrac{51}{17}=3\\ \Leftrightarrow\left\{{}\begin{matrix}x=21\\y=15\end{matrix}\right.\)
\(b,\) Áp dụng t/c dtsbn:
\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{x^2}{25}=\dfrac{y^2}{16}=\dfrac{x^2-y^2}{25-16}=\dfrac{1}{9}\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{9}\\y=\dfrac{4}{9}\end{matrix}\right.\)
\(c,\dfrac{x}{y}=\dfrac{2}{5}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{5}\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=k\Leftrightarrow x=2k;y=5k\)
\(xy=40\Leftrightarrow10k^2=40\Leftrightarrow k^2=4\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4;y=10\\x=-4;y=-10\end{matrix}\right.\)
