Question

已知:a+b+c=abc≠0求(1-b2)(1-c2)/bc+(1-a2)(1-c2)/ac+(1

Answer 1

原式=[(1-a^2)(1-b^2)c+(1-b^2)(1-c^2)a+(1-c^2)(1-a^2)b]/abc =[(c-a^2c-b^2c+a^2b^2c)+(a-ab^2-b^2c+ab^2c^2)+(b-bc^2-a^2b+a^2bc^2)]/abc =[(c-a^2c-b^2c+ab(a+b+c))+(a-ab^2-b^2c+bc(a+b+c))+(b-bc^2-a^2b+ac(a+b+c))]/abc =[(c-a^2c-b^2c+a^2b+ab^2+abc)+(a-ab^2-b^2c+abc+b^2c+bc^2)+(b-bc^2-a^2b+a^2c+abc+ac^2)]/abc =(c+a+b+3abc)/abc =4abc/abc =4