Đặt 2/3x = 3/4y = 4/5z= a
=> x= 3a/2; y= 4a/3; z= 5a/4
có x^2 + y^2 - 2z^2 = 520
=> (3a/2)^2+ (4a/3)^2- 2(5a/4)^2=520
<=> 9a^2/4+ 16a^2/9- 50a^2/16=520
<=> a^2( 9/4+ 16/9- 50/16)= 520
<=> 65/72. a^2= 520
<=> a^2= 576
TH1: a= 24
=> x= 36; y= 32; z= 30
TH2: a=-24
=> x= -36; y= -32; z=-30