求解a(1/b+1/c+1/d)+b(1/a+1/c+1/d)+c(1/b+1/a+1/d)+d(1/b+1/c+1/a)的解.条件是a+b+c+d=0
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求解a(1/b+1/c+1/d)+b(1/a+1/c+1/d)+c(1/b+1/a+1/d)+d(1/b+1/c+1/a)的解.条件是a+b+c+d=0
求解a(1/b+1/c+1/d)+b(1/a+1/c+1/d)+c(1/b+1/a+1/d)+d(1/b+1/c+1/a)的解.条件是a+b+c+d=0
求解a(1/b+1/c+1/d)+b(1/a+1/c+1/d)+c(1/b+1/a+1/d)+d(1/b+1/c+1/a)的解.条件是a+b+c+d=0
a(1/b+1/c+1/d)+b(1/a+1/c+1/d)+c(1/b+1/a+1/d)+d(1/b+1/c+1/a)
=(b+c+d)/a+(a+c+d)/b+(a+b+d)/c+(a+b+c)/d
=-a/a-b/b-c/c-d/d
=-1-1-1-1
=-4
560
a=-1 b=1 c=-1 d=1
=[(a^2)cd+(a^2)bd+(a^2)bc+(b^2)cd+a(b^2)d+a(b^2)c+a(c^2)d+b(c^2)d+ab(c^2)+ac(d^2)+ab(d^2)+bc(d^2)]/abcd
=[acd(a+c+d)+abd(a+b+d)+bcd(b+c+d)+abc(a+b+c)]/abcd
因为:a+b+c+d=0
所以:a+c+d=-b, a+b+d=...
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=[(a^2)cd+(a^2)bd+(a^2)bc+(b^2)cd+a(b^2)d+a(b^2)c+a(c^2)d+b(c^2)d+ab(c^2)+ac(d^2)+ab(d^2)+bc(d^2)]/abcd
=[acd(a+c+d)+abd(a+b+d)+bcd(b+c+d)+abc(a+b+c)]/abcd
因为:a+b+c+d=0
所以:a+c+d=-b, a+b+d=-c, b+c+d=-a, a+b+c=-d 带入上面代数式得到
=[acd*(-b)+abd*(-c)+bcd*(-a)+abc*(-d)]/abcd
=-4abcd/abcd
=-4
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