1.已知 x/a-b=y/b-c=z/c-a,若a不等于b不等于c,求x+y+z.
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1.已知 x/a-b=y/b-c=z/c-a,若a不等于b不等于c,求x+y+z.
2.已知a+b+c=0,求a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)的值.
3.化简:[(x+2)/(x^2+2x)-(x-1)/(x^2--4x+4)]/(x-4/x)*(2-x)^2
4.化简;(1/2x+6)+(1/3-x)+[x/2(x^2-9)
3.[(x+2)/(x^2+2x) - (x-1)/(x^2--4x+4)] /(x-4/x) *(2-x)^2
全隔开啦
2.已知a+b+c=0,求a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)的值.
3.化简:[(x+2)/(x^2+2x)-(x-1)/(x^2--4x+4)]/(x-4/x)*(2-x)^2
4.化简;(1/2x+6)+(1/3-x)+[x/2(x^2-9)
3.[(x+2)/(x^2+2x) - (x-1)/(x^2--4x+4)] /(x-4/x) *(2-x)^2
全隔开啦
1.设 x/a-b=y/b-c=z/c-a=k,
x=ka-kb,
y=kb-kc,
z=kc-ka,
三式相加,
x+y+z=ka-kb+kc-ka+kc-ka=0,
x+y+z=0
2,
a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)
=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
3,?
4.
(1/2x+6)+(1/3-x)+[x/2(x^2-9)]
=1/2(x+3)-1/(x-3)+x/2(x^2-9)
=[(x-3)-(x+3)]/(x^2-9)+x/2(x^2-9)
=(x-12)/2(x^2-9)
x=ka-kb,
y=kb-kc,
z=kc-ka,
三式相加,
x+y+z=ka-kb+kc-ka+kc-ka=0,
x+y+z=0
2,
a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)
=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
3,?
4.
(1/2x+6)+(1/3-x)+[x/2(x^2-9)]
=1/2(x+3)-1/(x-3)+x/2(x^2-9)
=[(x-3)-(x+3)]/(x^2-9)+x/2(x^2-9)
=(x-12)/2(x^2-9)
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