3道初一代数题1/根号3+1+1/根号5+根号3+1/根号7+根号5……1/根号2n-1+根号2n+1=?

1个回答

  • 1/(根号3+1)+1/(根号5+根号3)+1/(根号7+根号5)+……+1/(根号2n-1+根号2n+1)

    =(√3-1)/(√3+1)(√3-1)+(√5-√3)/(√5+√3)(√5-√3)+(√7-√5)/(√7+√5)(√7-√5)+.+[√(2n+1)-√(2n-1)]/[

    √(2n+1)+√(2n-1)](√(2n+1)-√(2n-1)]

    =(√3-1)/2+(√5-√3)/2+(√7-√5)/2+.+[√(2n+1)-√(2n-1)]/2

    =[√3-1+√5-√3+√7-√5+.+√(2n-1)+√(2n+1)-√(2n-1)]/2

    =[-1+√(2n+1)]/2

    (x-y)√[1/(y-x)]+1/3 4√(x²-2xy+y²)

    =(x-y)√[(y-x)/(y-x)²]+1/3 4√(x-y)²

    当x>y

    原式=(x-y)*1/(x-y)√(x-y)+1/3√(x-y)

    =√(x-y)+1/3√(x-y)

    =4/3√(x-y)

    当 x=y

    原式=0

    当 x<y

    原式=(x-y)*1/(y-x)√(y-x)+1/3√(y-x)

    =-√(y-x)+1/3√(y-x)

    =-2/3√(y-x)

    ³√(-14)^﹣9

    =³√(-1/14^9)

    =-1/14³

    =-1/2744

    -³√(-0.1)^-9

    =-³√(-1/0.1)^9

    =-³√(-10)^9

    =-(-10)³

    =-(-1000)

    =1000