Question

已知4x-3y-6z=0 x+2y-7z=0 （xyz不等于0）求(5x2+2y2-z2)除以（2x2-3y2-10z)的值

X后面的2的平方

Answer 1

解：4x-3y-6z=0(1)
x+2y-7z=0 (2)
(1)×2+(2)×3可得 2(4x-3y-6z)+3(x+2y-7z)=0整理得z=x/3
(1)×7-(2)×6可得 7(4x-3y-6z)-6(x+2y-7z)=0整理得y=2x/3
(5x²+2y²-z²)÷(2x²-3y²-10z²)
=[5x²+2(2x/3)²-(x/3)²]÷[2x²-3(2x/3)²-10(x/3)²]
=(5x²+8x²/9-x²/9)÷(2x²-4x²/3-10x²/9)
=52x²/9÷(-4x²/9)
=-13
(5x2+2y2-z2)除以（2x2-3y2-10z)中10z应该是10z²吧，否则约不掉未知数。

Answer 2

4x-3y-6z=0 x+2y-7z=0
所以，x=3z，y=2z

原式=[5(3z)^2+2(2z)^2-z^2]/[2(3z)^2-3(2z)^2-10z^2]
=(45+8-1)/(18-12-10)
=-13