Question

已知a2-b2=0,且ab≠0,求分式a2-3ab+5b2/3a2-2ab+4b2的值.

Answer 1

(a2-3ab+5b2) / (3a2-2ab+4b2 ) = [(a^2 - b^2) + 6b^2 - 3ab] / [3(a^2 - b^2) + 7b^2 - 2ab] = [6b^2 - 3ab] / [7b^2 - 2ab] = [6b - 3a] / [7b - 2a] (1) a+b = 0 ==> [6b - 3a] / [7b - 2a] = [6b+ 3b] / [7b + 2b] = 1 (2) a-b = 0 ==> [6b - 3a] / [7b - 2a] = [6b - 3b] / [7b - 2b] = 3/5 所以a2-3ab+5b2/3a2-2ab+4b2的值 为 1 或 3/5.