作业帮这几道数学题咋做啊 1.已知x+y+z=0,则1/（y²+z²-x²)+1/(z²+x²-y²)+1/(x²+y²-z²)= 2.若 3x-4y-z=0,2x+y-8z=0,则（x+y-z)/(x-y+z)= 3.已知a+b+c=0,且a,b,c均不为0,则a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/13 14:21:28
这几道数学题咋做啊 1.已知x+y+z=0,则1/（y²+z²-x²)+1/(z²+x²-y²)+1/(x²+y²-z²)= 2.若 3x-4y-z=0,2x+y-8z=0,则（x+y-z)/(x-y+z)= 3.已知a+b+c=0,且a,b,c均不为0,则a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/
这几道数学题咋做啊
1.已知x+y+z=0,则1/（y²+z²-x²)+1/(z²+x²-y²)+1/(x²+y²-z²)= 2.若 3x-4y-z=0,2x+y-8z=0,则（x+y-z)/(x-y+z)= 3.已知a+b+c=0,且a,b,c均不为0,则a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=
这几道数学题咋做啊 1.已知x+y+z=0,则1/（y²+z²-x²)+1/(z²+x²-y²)+1/(x²+y²-z²)= 2.若 3x-4y-z=0,2x+y-8z=0,则（x+y-z)/(x-y+z)= 3.已知a+b+c=0,且a,b,c均不为0,则a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/
1.∵x+y+z=0
∴y²+z²-x²=y²+z²-(y+z)²= -2yz
同理z²+x²-y²= -2zx x²+y²-z²= -2xy
原式=1/-2yz+ 1/-2zx+ 1/ -2xy=(通分）-（x+y+z)/ -2xyz =o (分子x+y+z=0）
2.3x-4y-z=0 （1） 2x+y-8z=0（2）
（1）+（2）×4得 11x-33z=0 所以x=3z z=1/3 x
(1)×8 - (2)得 22x-33y=0 ∴2x=3y y=2/3x
带入原式
原式=(x+2/3x—1/3 x）/(x -2/3x+1/3 x)=2
3.原式=a(1/b+1/c+1/a)+b(1/c+1/a+1/b)+c(1/a+1/b+1/c)-3 (每项加一个1）
=(1/b+1/c+1/a)（a+b+c) -3 (合并同类项 将三项合并）
= -3
做完啊...详细吧.
1、-x=y+z x^2=y^2+z^2+2yz y^2+z^2-x^2=-2yz
同理，z^2+x^2-y^2=-2zx，x^2+y^2-z^2=-2xy
原式=-1/2yz-1/2zx-1/2xy
=-(x+y+z)/2xyz
=0
2、z=3x-4y y=8z-2x=8(3x-4y)-2x 33y=22x y=2x/3 ...
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1、-x=y+z x^2=y^2+z^2+2yz y^2+z^2-x^2=-2yz
同理，z^2+x^2-y^2=-2zx，x^2+y^2-z^2=-2xy
原式=-1/2yz-1/2zx-1/2xy
=-(x+y+z)/2xyz
=0
2、z=3x-4y y=8z-2x=8(3x-4y)-2x 33y=22x y=2x/3 z=3x-8x/3=x/3
原式=(x+2x/3-x/3)/(x-2x/3+x/3)=(4/3)/(2/3)=2
3、原式=a/b+a/c+b/c+b/a+c/a+c/b
=(a+c)/b+(a+b)/c+(b+c)/a
=-1-1-1
=-3
收起