Question

1.已知x=2-根号3，求（x-1分之1-2x+x²） -（x²-x分之根号x²-2x+1）的值

2.已知a+b=5,a²+b²=13，化简并求b倍根号b分之a +a倍根号a分之b的值

谢谢！

Answer 1

x-1=1-3^(1/2)<0.
(x^2-2x+1)^(1/2)=[(x-1)^2]^(1/2)=|x-1|=1-x
(1-2x+x^2)/(x-1) - (x^2-2x+1)^(1/2)/(x^2-x) = (x-1)^2/(x-1) - (1-x)/[x(x-1)]
=x-1 + (x-1)/[x(x-1)]
=x-1+1/x
=(x^2-x+1)/x

2ab = (a+b)^2 - (a^2+b^2) = 5^2 - 13 = 12,
ab = 6.
b*(a/b)^(1/2) + a*(b/a)^(1/2) = (ab)^(1/2) + (ab)^(1/2) = 2(ab)^(1/2) = 2*(6)^(1/2)