Sửa đề: \(x^2-y^2+2z^2=108\)
Đặt x/2=y/3=z/5=k
=>x=2k; y=3k; z=5k
Ta có: \(x^2-y^2+2z^2=108\)
\(\Leftrightarrow4k^2-9k^2+50k^2=108\)
=>45k2=108
=>k2=12/5
TH1: \(k=\dfrac{2\sqrt{3}}{\sqrt{5}}\)
=>\(x=\dfrac{4\sqrt{3}}{\sqrt{5}};y=\dfrac{6\sqrt{3}}{\sqrt{5}};z=2\sqrt{15}\)
TH2: \(k=-\dfrac{2\sqrt{3}}{\sqrt{5}}\)
=>\(x=-\dfrac{4\sqrt{3}}{\sqrt{5}};y=-\dfrac{6\sqrt{3}}{\sqrt{5}};z=-2\sqrt{15}\)