若2x-3y-z=0,x+3y-14z=0,且z≠0,则x²+3xy/y²+z²的值为多少?
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若2x-3y-z=0,x+3y-14z=0,且z≠0,则x²+3xy/y²+z²的值为多少?
若2x-3y-z=0,x+3y-14z=0,且z≠0,则x²+3xy/y²+z²的值为多少?
若2x-3y-z=0,x+3y-14z=0,且z≠0,则x²+3xy/y²+z²的值为多少?
令2x-3y-z=0为方程【1】,x+3y-14z=0为方程【2】
【1】+【2】得出x=5z
2【2】-【1】得出y=3z
将x与y的值代入x²+3xy/y²+z²,根据z≠0,化简得到 7.
(1)+(2)可得X=5Z (2)*2-(1)S可得Y=3Z
x²+3xy/y²+z²=X(X+3Y)/(y²+z²)=X*14Z/(y²+z²)=5Z*14Z/(9z²+z²)=7
答案为7.步骤有前两个公式得出3Y=2X-Z=14Z-X,可得出X=5Z,Y=3Z,,带入最后的式子得出答案7
2x-3y-z=0
2x-3y=z 1
1式代入x+3y-14z=0得
x+3y-14(2x-3y)=0
x+3y-28x+42y=0
45y=27x
5y=3x
x/5=y/3 设=k
x=5k y=3k 代入1式得
z=2*5k-3*3k=k
x^2+3xy/y^2+z^2
=(5k)^2+3*5k*3k/(3k)^2+k^2
=25k^2+45k^2/9k^2+k^2
=70k^2/10k^2
=7
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