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
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