设√5+1/(√5-1)的整数部分为a,小数部分为b,求a^2+1/2*ab+b^2的值
问题描述：
设√5+1/(√5-1)的整数部分为a,小数部分为b,求a^2+1/2*ab+b^2的值
答
a+b=√5+1/(√5-1);
a^2+1/2*ab+b^2 = (a+b)^2 =[√5+1/(√5-1)]^2 = (3+√5)/(3-√5);
看错了，呵呵，作废
答
[5^(1/2) + 1]/[5^(1/2) - 1]
= [5^(1/2) + 1]^2/[5 - 1]
= [6 + 2*5^(1/2)]/4
= [3 + 5^(1/2)]/2
= 2 + [5^(1/2) - 1]/2
5^(1/2) - 1 > 0.
0 = [5 - 1]/{2[5^(1/2) + 1]}
= 2/[5^(1/2) + 1]
= 2/3 所以,
a = 2,
b = [5^(1/2) - 1]/2
a^2 + 1/2*ab + b^2
= 4 + 1/2*2*[5^(1/2) - 1]/2 + {[5^(1/2) - 1]/2}^2
= 4 + [5^(1/2) - 1]/2 + [6 - 2*5^(1/2)]/4
= 4 + [5^(1/2) - 1]/2 + [3 - 5^(1/2)]/2
= 4 - 1/2 + 3/2
= 5