Question

已知根号x-6分之a-x=根号x-6分之根号a-x,且x为偶数,求(1+x)倍根号x2-5x+4分之x2-1

Answer 1

根号x-6分之a-x=根号x-6分之根号a-x

则(a-x)^2=a-x

解得 x=a 或x=a-1

(1+x)倍根号x2-5x+4分之x2-1

=(x-1)(x+1)/(x+1)√(x-4)(x-1)

=√[(x-1)/(x-4)]

代入x=a 原式=√[(a-1)/(a-4)]

代入x=a-1 原式=√[(a-2)/(a-5)]