2道二次根式（3根号2+根号3-根号5）^2-(3根号2-根号3+根号5）^2 （1+根号2）^2(1+根号3）^2(1-根号2）^2(1-根号3）^2

Question

2道二次根式
（3根号2+根号3-根号5）^2-(3根号2-根号3+根号5）^2 （1+根号2）^2(1+根号3）^2(1-根号2）^2(1-根号3）^2

Answer 1

(3√2+√3-√5)^2-(3√2-√3+√5)^2
=[(3√2+√3-√5)-(3√2-√3+√5)][(3√2+√3-√5)+(3√2-√3+√5)]
=(3√2+√3-√5-3√2+√3-√5)(3√2+√3-√5+3√2-√3+√5)
=(2√3-2√5)*6√2
=2(√3-√5)*6√2
=12(√3-√5)*√2
=12√6-12√10
（1+√2)^2(1+√3)^2(1-√2)^2(1-√3）^2
=(1+√2)^2(1-√2)^2(1-√3)^2(1+√3)^2
=[(1+√2)^2(1-√2)^2][(1-√3)^2(1+√3)^2]
=[(1+√2)(1-√2)]^2[(1-√3)(1+√3)]^2
=[1-2]^2*[1-3]^2
=1*4
=4