已知4x-3y-3z=0,x-3y-z=0.求:(1)x:y:z的值(2)(xy+2yz)/(x^2+y^2-z^2)的值
来源:百度知道 编辑:UC知道 时间:2024/06/03 20:12:42
由4x-3y-3z=0,x-3y-z=0 得:
x=2z/3,y=-z/9
x:y:z=2z/3:(-z/9):z=6:(-1):9
(xy+2yz)/(x^2+y^2-z^2)
=(-2z^2/27-2z^2/9)/(4z^2/9+z^2/81-z^2)
=(-8/27)/(-46/81)
=12/23
已知4x-3y-3z=0, a式
x-3y-z=0. b式
将两式联立解方程组
a式-b式消y,得:
x=2/3*z,代入b式,得y=-1/9*z
x:y:z=2/3*z:-1/9*z:z=6:-1:9
(2)(xy+2yz)/(x^2+y^2-z^2)=(2/3*z*-1/9*z+2*-1/9*z*z)/[(2/3*z)^2+(-1/9*z)^2-z^2]
=(-8/27)/(-44/81)
=135/44
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