求20道二次根式计算题,

1个回答

  • 1.X(2-√2)=(√2-2)

    X(2-√2)=(√2-2)

    X(2-√2)=-1(2-√2)

    x=1

    2.(√5+√2)²-(√5-√2)²

    =(√5+√2+√5-√2)(√5+√2-√5+√2)

    =2√5*2√2

    =4√10

    3.√8+3√(1/3)-1/(√2)+(√3)/2

    =2√2+3*(√3)/3-(√2)/2+(√3)/2

    =2√2+√3-(√2)/2+(√3)/2

    =(3/2)√3+(3/2)√2

    4.(√3+√2+√5)(√3-√2-√5)

    =(√3+√2+√5)[√3-(√2+√5)]

    =(√3)²-(√2+√5)²

    =3-(2+5+2√10)

    =3-7-2√10

    =-4-2√10

    5.√8-2√32+√50

    =5*3√2-2*4√2+5√2

    =√2(15-8+5)

    =12√2

    6.√6-√3/2-√2/3

    =√6-√6/2-√6/3

    =√6/6

    7.(√45+√27)-(√4/3+√125)

    =(3√5+3√3)-(2√3/3+5√5)

    =-2√5+7√5/3

    8.(√4a-√50b)-2(√b/2+√9a)

    =(2√a-5√2b)-2(√2b/2+3√a)

    =-4√a-6√2b

    9.(√32-3√3)(4√2+√27)

    =(4√2-3√3)(4√2+3√3)

    =(4√2)^2-(3√3)^2

    =32-27

    =5

    10.1+√2-√3)(1-√2+√3)

    =[1+(√2-√3)][1-(√2-√3)]

    =1-(√2-√3)^2

    =1-(2+3+2√6)

    =-4-2√6

    美女,10道够不够啊...