这几道数学题咋做啊 1.已知x+y+z=0,则1/(y²+z²-x²)+1/(z²

问题描述:

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)=
1个回答 分类:数学 2014-11-23

问题解答:

我来补答
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.
 
 
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