\(x+y-z=2\Rightarrow z=x+y-2\)
\(3x^2+2y^2-z^2=13\)
\(\Leftrightarrow3x^2+2y^2-\left(x+y-2\right)^2=13\)
\(\Leftrightarrow2x^2+y^2-2xy+4x+4y=17\)
\(\Leftrightarrow x^2+y^2+4-2xy-4x+4y+x^2+8x+16=37\)
\(\Leftrightarrow\left(x-y-2\right)^2+\left(x+4\right)^2=37=1^2+6^2\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-y-2\right)^2=1\\\left(x+4\right)^2=6^2\end{matrix}\right.\) (do \(x\) nguyên dương nên chỉ có TH này)
\(\Rightarrow\left\{{}\begin{matrix}x-y-2=1\\x+4=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\) (loại)
Hoặc \(\left\{{}\begin{matrix}x-y-2=-1\\x+4=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Câu 2:
\(a^2+b^2=c^2\Leftrightarrow\left(a+b\right)^2-2ab=c^2\)
\(\Leftrightarrow2ab=\left(a+b\right)^2-c^2=\left(a+b-c\right)\left(a+b+c\right)\) (1)
\(\Rightarrow2ab⋮\left(a+b+c\right)\)
- Nếu \(a+b+c\) lẻ \(\Rightarrow2⋮̸\left(a+b+c\right)\Rightarrow ab⋮\left(a+b+c\right)\)
- Nếu \(a+b+c\) chẵn, ta có \(\left(a+b+c\right)+\left(a+b-c\right)=2\left(a+b\right)\) chẵn
\(\Rightarrow a+b-c=2\left(a+b\right)-\left(a+b+c\right)\) là hiệu của 2 số chẵn \(\Rightarrow\) là số chẵn
\(\Rightarrow a+b-c=2k\) thay vào (1) ta được
\(\Rightarrow2k\left(a+b+c\right)=2ab\) \(\Rightarrow ab=k\left(a+b+c\right)\Rightarrow ab⋮\left(a+b+c\right)\)
