\(\frac{x}{2}=\frac{y}{5}\Rightarrow x=2k\); \(y=5k\)
Ta có : \(2k.5k=90\Rightarrow10k^2=90\Rightarrow k^2=9\Rightarrow\orbr{\begin{cases}k=3\\k=-3\end{cases}}\)
Với \(k=3\Rightarrow x=2.3=6\); \(y=5.3=15\)
Với \(k=-3\Rightarrow x=2.-3=-6\); \(y=5.-3=-15\)
Vậy ....
Đặt :
\(\frac{x}{2}=\frac{y}{5}=k\Leftrightarrow x=2k;y=5k\)
Thay \(x=2k;y=5k\) vào \(x.y=90\) Ta có :
\(2k.5k=90\)
\(\Leftrightarrow10.k^2=90\)
\(\Leftrightarrow k^2=9\)
\(\Leftrightarrow k=3\)
+) \(k=3\Leftrightarrow\hept{\begin{cases}x=2k=2.3=6\\y=5k=5.3=15\end{cases}}\)
Vậy .................
\(\frac{x}{2}=\frac{y}{5}\Rightarrow x=2k;y=5k\)
Ta có :\(2k.5k=90\)
\(\Rightarrow10.k^2=90\rightarrow k^2=9\rightarrow k=3;k=-3\)
\(k=3\Rightarrow x=2.3=6\)
\(y=5.3=15\)
\(k=-3\Rightarrow x=2.\left(-3\right)=-6\)
\(y=5.\left(-3\right)=-15\)
Vậy ....
\(\frac{x}{2}=\frac{y}{5}\Rightarrow x=2k;y=5k\)
Do x.y=90
\(\Rightarrow2k.5k=90\Rightarrow10k^2=90\Rightarrow k^2=9\Rightarrow\orbr{\begin{cases}k=3\\k=-3\end{cases}}\)
\(\orbr{\begin{cases}x=2.3=6;y=5.3=15\\x=2.\left(-3\right)=-6;y=5.\left(-3\right)=-15\end{cases}}\)
Vậy ....... :))
#Nguyễn Hà Thảo Vy:copy nhớ ghi nguồn nha b:)