数学题若x^2-4分之4x=x+2分之
x^2-4分之4x=x+2分之a减去x-2分之b
[4x/(x^2-4)]=[a/(x+2)]-[(b/(x-2)]=====>
4x=(a-b)x-2a-2b =======>
a-b=4
a+b=0======>
a=2 b=-2========>
a^2+b^2=8
x(1/y+1/z)+y(1/x+1/z）+z(1/x+1/y)=(x+z)/y+(x+y)/z+(y+z)/x
(原式变形)
=-1-1-1=-3
(1)
y+1/x+1+x+1/y+1=（y+1)/(x+1)+(x+1)/(y+1)
=(xx+yy+2xy-2xy+2x+2y+2)/(xy+x+y+1)
=-34/1...全部
x^2-4分之4x=x+2分之a减去x-2分之b
[4x/(x^2-4)]=[a/(x+2)]-[(b/(x-2)]=====>
4x=(a-b)x-2a-2b =======>
a-b=4
a+b=0======>
a=2 b=-2========>
a^2+b^2=8
x(1/y+1/z)+y(1/x+1/z）+z(1/x+1/y)=(x+z)/y+(x+y)/z+(y+z)/x
(原式变形)
=-1-1-1=-3
(1)
y+1/x+1+x+1/y+1=（y+1)/(x+1)+(x+1)/(y+1)
=(xx+yy+2xy-2xy+2x+2y+2)/(xy+x+y+1)
=-34/15
(2)
解方程a^2+4a-12=0
得：
x=a1=2
y=a2=-6
原式=-5/3-3/5=-34/15
。
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