Question

(1)2-l1-2√2l-(√6-3)^0 l是绝对值 (2)3√40-√2/5-2√1/10 (3)(√5-√5分之1)^2

Answer 1

（1）2-l1-2√2l-(√6-3)^0
=2-(2√2-1)-1
=2-2√2+1-1
=2-2√2

（2）3√40-√(2/5)-2√(1/10)
=6√10-(1/5)√10-(1/5)√10
=(6-1/5-1/5)√10
=(28/5)√10

（3）(√5-√5分之1)^2
=[√5-(1/5)√5]²
=[(1-1/5)√5]²
=[(4/5)√5]²
=16/25x5
=16/5

追问

(1)√18+1/2√12-√9
(2)6√1/2+4√0.75-3√2
(3)(√3-2)^2002*(√3+2)^2003

Answer 2

(1)2-l1-2√2l-(√6-3)^0 l是绝对值
=2-2√2+1-1
=2-2√2
(2)3√40-√2/5-2√1/10
=6√10-1/5√10-1/5√10
=28/5√10
(3)(√5-√5分之1)^2
=5-2+1/5
=3.2

追问

(1)√18+1/2√12-√9
(2)6√1/2+4√0.75-3√2
(3)(√3-2)^2002*(√3+2)^2003

追答

(1)√18+1/2√12-√9
=3√2+√3-2
(2)6√1/2+4√0.75-3√2
=3√2+2√3-3√2
=2√3
(3)(√3-2)^2002*(√3+2)^2003
=【(√3-2)*(√3+2)】^2002*(√3+2)
=【3-2】^2002*(√3+2)
=√3+2

追问

√5-√2/5-2-√5的绝对值要过程
3(√10+1)^0+√1/3*√12要过程
(-3)^-1+√8+√2-2/3的绝对值要过程

追答

算了，你自己做吧

Answer 3

• 2-l1-2√2l-(√6-3)^0 

• =2-（2√2-1）-0

• =2-2√2l-1

• =1-2√2l