Đặt \(\frac{x}{5}=\frac{y}{7}=k\Rightarrow x=5k;y=7k\)
Thay \(x=5k;y=7k\) ta có:
\(5k\times7k=875\)
\(\Leftrightarrow k^2=875\div\left(5\times7\right)\)
\(\Leftrightarrow k^2=25\)
\(\Rightarrow k=\pm5\)
TH1: \(k=5\)
\(\Rightarrow\hept{\begin{cases}x=5\times5=25\\y=7\times5=35\end{cases}}\)
TH2: \(k=-5\)
\(\Rightarrow\hept{\begin{cases}x=5\times\left(-5\right)=-25\\y=7\times\left(-5\right)=-35\end{cases}}\)
Vậy ...
Đặt :
\(\frac{x}{5}=\frac{y}{7}=k\) \(\Leftrightarrow\hept{\begin{cases}x=5k\\y=7k\end{cases}}\)
Mà \(xy=875\)
\(\Leftrightarrow5k.7k=875\)
\(\Leftrightarrow35k^2=875\)
\(k^2=25\)
\(\Leftrightarrow\orbr{\begin{cases}k=5\\k=-5\end{cases}}\)
+) Với \(k=5\Leftrightarrow\hept{\begin{cases}x=25\\y=35\end{cases}}\)
+) Với \(k=-5\Leftrightarrow\orbr{\begin{cases}x=-25\\y=-35\end{cases}}\)
Ta có: \(x.y=875\Rightarrow x=875:y_{\left(1\right)}\)
Thay (1) vào \(\frac{x}{5}=\frac{y}{7}\), ta được:
\(\frac{875:y}{5}=\frac{y}{7}\Rightarrow\frac{175}{y}=\frac{y}{7}\)
\(\Leftrightarrow175.7=y.y\)
\(\Leftrightarrow1225=y^2\)
\(\Leftrightarrow y=\pm35\)
\(y=35\Rightarrow x=875:y=875:35=25\)
\(y=-35\Rightarrow x=875:\left(-35\right)=-25\)
Vậy \(y=35;x=25\)
\(y=-35;x=-25\)
Tớ nek, hi bn
Đặt \(\frac{x}{5}=\frac{y}{7}=k\) \(\Rightarrow\)\(x=5k;\)\(y=7k\)
Ta có: \(x.y=875\)
\(\Rightarrow\)\(5k.7k=875\)
\(\Rightarrow\)\(k^2=25\)
\(\Rightarrow\)\(k=\pm5\)
\(k=5\)\(\Rightarrow\)\(x=25;\)\(y=35\)\(k=-5\)\(\Rightarrow\)\(x=-25;\)\(y=-35\)
