已知a、b、c为不等于零的实数,且a+b+c=0,求a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)的值
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已知a、b、c为不等于零的实数,且a+b+c=0,求a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)的值
已知a、b、c为不等于零的实数,且a+b+c=0,求a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)的值
已知a、b、c为不等于零的实数,且a+b+c=0,求a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)的值
因为a+b+c=0 所以 a=-b-c
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
=(b+c)a+(a+c)/b+(a+b)/c
把 a=-b-c 带入上面式子得:=(b+c)/(-b-c) + (-b-c+c)/b+(-b-c+b)/c
=-1+(-1)(-1)
=-3
∵a+b+c=0
∴ a+b=-c, a+c=-b, b+c=-a
a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)
=a(b+c)/bc+b(a+c)+c(a+b)/ab
=-a²/bc-b²/ac-c²/ab
=-(a³+b³+...
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∵a+b+c=0
∴ a+b=-c, a+c=-b, b+c=-a
a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)
=a(b+c)/bc+b(a+c)+c(a+b)/ab
=-a²/bc-b²/ac-c²/ab
=-(a³+b³+c³)/abc
=-[(a+b)³+c³-3ab(a+b)]/abc
=-{(a+b+c)[(a+b)²-(a+b)c+c²]+3abc}/abc
=-3abc/abc
=-3.
收起
因为,a+b+c=0
所以,a=-b-c
所以,原式=(-b-c)(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=-1-b/c-c/b-1+b/c+b/a+c/a+c/b
=-2+(b+c)/a=-2+(-a)/a=-3
= a/b+a/c+b/c+b/a+c/a+c/b
=(a+c)/b+(b+a)/c+(b+c)/a
=-1-1-1
=-3
原式加3,a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b) + 3
= a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b) + a/a + b/b + c/c
= (a+b+c)(1/a+1/b+1/c) = 0
原式为-3