Question

当a=2+根号6,b=2-根号6时,求代数式（a-b+4ab/a-b)*(a+b-4ab/a+b)的值.

Answer 1

由已知,a = 2 + √6,b = 2 –√6,所以a – b = 2√6 ；而且a + b = 4 ；而且a*b = (2 + √6)*(2 –√6) = 2 2 –(√6) 2 = 4 – 6 = -2 ； 原式= [(a – b + 4ab)/(a – b)]*[(a + b – 4ab)/(a + b)] = [2√6 + 4*(-2)]/(2√6)*[(4– 4*(-2)]/4 = (2√6 – 8)/(2√6)*(12/4) = 3(√6 – 4)/(√6) = √6(√6 – 4)/2 = (6 – 4√6)/2= 3– 2√6 .