Question

请教一道二次根式的数学题！！！！

√3-√2+1/√2+1-2/√3+1
(3-√7/2)^2+√7+3/√7-3

√3/(5-√7）- （2√3-5√7）^2

（2√3-2) （√6+√2）+（3√3+5）/(3√3-5)

Answer 1

√3-√2+1/(√2+1)-2/(√3+1)
=√3-√2+(√2-1)/(√2+1)(√2-1)-2(√3-1)/(√3+1)(√3-1)
=√3-√2+√2-1-(√3-1)
=0
[(3-√7)/2]^2+(√7+3)/(√7-3)
=[(3-√7)/2]^2+(√7+3)(√7+3)/(√7-3)(√7+3)
=(3-√7)^2/4+(√7+3)^2/4
=(16-6√7)/4+(16+6√7)/4
=(16+16)/4
=8
√3/(5-√7)- (2√3-5√7)^2
=√3(5+√7)/(5-√7)(5+√7)- (2√3-5√7)^2
=(5√3+√21)/18-(187-20√21)
=5√3/18+36√21/18+187

(2√3-2)(√6+√2)+(3√3+5)/(3√3-5)
=2(√3-1)*√2(√3+1)+(3√3+5)^2/2
=4√2+26+15√3
=15√3+4√2+26

Answer 2

是否会有括号？这不是直接计算么？

追问

√3-√2+1/√2+1-2/√3+1 没有括号