Question

已知a=2+根号3分之1,求a+2分之a²-4-a²-a分之根号a²-2a+1的值

Answer 1

a=1/(2+√3)
=(2-√3)/[(2-√3)(2+√3)]
=(2-√3)/(2²-√3²)
=(2-√3)/(4-3)
=2-√3
1/a=2+√3

原式=[(a²-4)/(a+2)]-[√(a²-2a+1)]/(a²-a)
=[(a-2)(a+2)/(a+2)]-[√(a-1)²]/[a(a-1)]
=a-2-|a-1|/[a(a-1)]
∵a=2-√3﹥0
∴原式=a-2-(a-1)/[a(a-1)]
=a-2-1/a
把a=2-√3, 1/a=2+√3代入原式
原式=2-√3-2-(2+√3)
=-√3-2-√3
=-2-2√3

Answer 2

a-1分之a的平方-1减a的平方-a分之根号a的平方-2a+1
=(a-1)(a+1)/(a-1)-(a-1)²/[a(a-1)]
=a+1-(a-1)/a
=(a²+a-a+1)/a
=(a²+1)/a
=a+1/a
=2-√3+1/(2-√3)
=2-√3+2+√3
=4

Answer 3

请查收

Answer 4

你问的不清楚，尤其是分号那