mk biết có mỗi cách thôi, làm xằng vậy =))
Ta có :
\(5x=3y\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{5}\)
Đặt :
\(\dfrac{x}{3}=\dfrac{y}{5}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\)\(\left(1\right)\)
Thay \(\left(1\right)\) vào \(xy=375\) ta có :
\(3k.5k=375\)
\(\Leftrightarrow15k^2=375\)
\(\Leftrightarrow\left[{}\begin{matrix}k=25\\k=-25\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\)
+) \(k=5\Leftrightarrow\left[{}\begin{matrix}x=3.5=15\\y=5.5=25\end{matrix}\right.\)
+) \(k=-5\Leftrightarrow\left[{}\begin{matrix}x=3.\left(-5\right)=-15\\y=5\left(-5\right)=-25\end{matrix}\right.\)
Vậy ..
1 Cách nhé
\(5x=3y\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{5}\)
Đặt : \(\dfrac{x}{3}=\dfrac{y}{5}=k\)
Ta có : \(\dfrac{x}{3}=k\Leftrightarrow x=3k\)
\(\dfrac{y}{5}=k\Leftrightarrow y=5k\)
Mà : \(3k.5k=375\)
\(\Leftrightarrow15k^2=375\)
\(\Leftrightarrow k^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\)
Khi \(k=5\), thì : \(\left\{{}\begin{matrix}x=15\\y=25\end{matrix}\right.\)
Khi \(k=-5,\) thì : \(\left\{{}\begin{matrix}x=-15\\y=-25\end{matrix}\right.\)
Vậy ..........
Giải:
Ta có: \(5x=3y\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{5}\)
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\)
Mà \(xy=375\)
\(\Leftrightarrow3k.5k=375\)
\(\Leftrightarrow15k^2=375\)
\(\Leftrightarrow k^2=\dfrac{375}{15}=25\)
\(\Leftrightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3.5\\x=3.\left(-5\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-15\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y=5.5\\x=5.\left(-5\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=25\\x=-25\end{matrix}\right.\)
Vậy ...
Chúc bạn học tốt!
