已知:(a+1)(b+1)/(x+2) + (a-1)(b-1)/(x-2) = 2ab/x 求:b/a + a/b

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已知:(a+1)(b+1)/(x+2) + (a-1)(b-1)/(x-2) = 2ab/x 求:b/a + a/b
已知:(a+1)(b+1)/(x+2) + (a-1)(b-1)/(x-2) = 2ab/x
求:b/a + a/b

已知:(a+1)(b+1)/(x+2) + (a-1)(b-1)/(x-2) = 2ab/x 求:b/a + a/b
化简得:x2 -2(a+b)x +4ab = 0
1.假设△<0,△= [-2(a+b)]的平方 = …… = [2(a-b)]的平方
(……)的平方必定≥0
所以矛盾,假设不成立.
2.令x=0 or 2 or -2
(1)x=0,4ab=0,b/a + a/b无解
(2)x=2,ab-a-b+1=0,b/a + a/b = 2
(3)x=-2,ab+a+b+1=0,b/a + a/b = 2