\(x^2+y^2=z^2+65\Rightarrow x^2+y^2-z^2=65\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{z}{3}=\dfrac{x^2+y^2-z^2}{49+25-9}=\dfrac{65}{65}=1\)
\(\dfrac{x}{7}=1\Rightarrow x=7\\ \dfrac{y}{5}=1\Rightarrow y=5\\ \dfrac{z}{3}=1\Rightarrow z=3\)
x2+y2=z2+65
⇒x2+y2-z2=65
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x}{7}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{3}\)
\(\dfrac{x^{^2}}{49}\)=\(\dfrac{y^{^2}}{25}\)=\(\dfrac{z^{^2}}{9}\)=\(\dfrac{x^{^2}+y^{^2}-z^{^2}}{49+25-9}\)=\(\dfrac{65}{65}\)=1
x=7
y=5
z=3
Đặt \(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{z}{3}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=7k\\y=5k\\z=3k\end{matrix}\right.\)
Ta có: \(x^2+y^2=z^2+65\)
\(\Leftrightarrow65k^2=65\)
\(\Leftrightarrow k^2=1\)
Trường hợp 1: k=1
\(\Leftrightarrow\left\{{}\begin{matrix}x=7k=7\\y=5k=5\\z=3k=3\end{matrix}\right.\)
Trường hợp 2: k=-1
\(\Leftrightarrow\left\{{}\begin{matrix}x=7k=-7\\y=5k=-5\\z=3k=-3\end{matrix}\right.\)
