求几道奥数题1.若x^2+4y^2+9z^2=x^3+2y^3+3z^3=x^4+y^4+z^4求X,Y,Z及上述个式的
1个回答
x^2+4y^2+9z^2=x^4+y^4+z^4
x²(x²-1)+y²(y²-4)+z²(z²-9)=0
x²=1,y²=4,z²=9
x^2+4y^2+9z^2=x^3+2y^3+3z^3
x²(x-1)+2y²(y-2)+3z²(z-3)=0
所以x=1,y=2,z=3
剩下的自己代进去!
3
X+Y+Z=1
x^2+y^2+z^2+2xy+2xz+2yz=1
因为x+y^2+z^2=2
xy+xz+yz=-1/2
X^3+Y^3+Z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=1*(2+1/2)
所以
3-3xyz=5/2
xyz=1/6
4
a+b+c=0,a*a+b*b+c*c=1,
(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc
2(ab+bc+ac)=(a+b+c)^2-(a^2+b^2+c^2)=0-1=-1
ab+bc+ac=-1/2,
a^4+b^4+c^4
=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+a^2c^2)
=1-2[(ab+bc+ac)^2-2abc(a+b+c)]
=1-2[(-1/2)^2-0]
=1-1/2
=1/2
7
b^2+c^2-a^2=(b+c)^2-2bc-a^2=a^2--2bc-a^2=-2bc
同理c^2+a^2-b^2=-2ac
a^2+b^2-c^2=-2ab
所以原式=1/（-2bc)+1/(-2ac)+1/(-2ab)
=(a+b+c)/(-2abc)
=0
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