已知abc=1,a+b+c=2,a²+b²+c²=3,那么1/ab+c-1 + 1/bc+a-1 + 1/ca+b-1 的值急
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![已知abc=1,a+b+c=2,a²+b²+c²=3,那么1/ab+c-1 + 1/bc+a-1 + 1/ca+b-1 的值急](/uploads/image/z/1993315-67-5.jpg?t=%E5%B7%B2%E7%9F%A5abc%3D1%2Ca%2Bb%2Bc%3D2%2Ca%26sup2%3B%2Bb%26sup2%3B%2Bc%26sup2%3B%3D3%2C%E9%82%A3%E4%B9%881%2Fab%2Bc-1+%2B+1%2Fbc%2Ba-1+%2B+1%2Fca%2Bb-1+%E7%9A%84%E5%80%BC%E6%80%A5)
已知abc=1,a+b+c=2,a²+b²+c²=3,那么1/ab+c-1 + 1/bc+a-1 + 1/ca+b-1 的值急
已知abc=1,a+b+c=2,a²+b²+c²=3,那么1/ab+c-1 + 1/bc+a-1 + 1/ca+b-1 的值
急
已知abc=1,a+b+c=2,a²+b²+c²=3,那么1/ab+c-1 + 1/bc+a-1 + 1/ca+b-1 的值急
因为abc=1
所以1/ab=c 1/bc=a 1/ca=b
1/ab+c-1+1/bc+a-1+1/ca+b-1
=1/ab+1/ac+1/bc+a+b+c-3
=a+b+c+a+b+c-3
=2a+2b+2c-3
=2(a+b+c)-3
=2*2-3
=1
1 / ab + c - 1 + 1 / bc + a - 1 + 1 / ca + b - 1
=(c + abc² - 1 + a + a²bc - 1 + c + ab²c - 1) / abc
=(c(1 + abc) + a(1 + abc) + b(1 + abc) - 3abc) / abc
=((a + c + b)(1 + abc)) / abc
=(2 × 2) / 1
=4
原式=1/ab+1/bc+1/ac+(a+b+c)-3
=(a+b+c)/abc+2-3
=2/1+2-3
=1
1/ab+c-1 + 1/bc+a-1 + 1/ca+b-1= (1/ab+1/bc+1/ac)+c+b+a-3= (a+b+c)/abc+(a+b+c)-3=2/1+2-3=1