已知:4x-3y-6z=0,2x-4y-14z=0求(2x^2+3y^2+z^2)/(x^2+5y^2+7z^2) 的值.
来源:百度知道 编辑:UC知道 时间:2024/05/29 10:19:40
4x-3y-6z=0 (1)
2x-4y-14z=0 (2)
(1)-(2)*2
4x-3y-6z-4x+8y+28z=0
5y+22z=0
y=-22z/5
y^2=19.36z^2
(1)*4-(2)*3
16x-12y-24z-6x+12y+42z=0
10x+18z=0
x=-9z/5=3.24z^2
(2x^2+3y^2+z^2)/(x^2+5y^2+7z^2)
=(6.48z^2+58.08z^2+z^2)/(3.24z^2+96.8z^2+7z^2)
=65.56z^2/107.04z^2
=1639/2676
把Z看作常数解关于X和Y的方程得X=-9Z/5,Y=-22Z/5,然后代入上式即可求解
由4x-3y-6z=0,
2x-4y-14z=0
解得:x=-9z/5
y=-22z/5
(2x^2+3y^2+z^2)/(x^2+5y^2+7z^2)
=[2(-9z/5)^2+3(-22z/5)^2+z^2]/[(-9z/5)^2+5(-22z/5)^2+7z^2]
=(162+1452+25)/(81+2420+175)
=1639/2676
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