1/(1+√2)+1/(√2+√3)+...+1/(√9999+√10000)=99
1/(1+√2)+1/(√2+√3)+...+1/(√9999+√10000)=99
√(1-√2)^2+√(√2-√3)^2.+√(√99-√100)
√1/5=(),计算:√2/3-√3/2,√18-√8+√1/8,√25/2+√99-√18
/√1-√2/+/√2-√3/+…+/√99-√100/
计算:√4-√3/√12+√3-√2/√6+√2-√1/√2=?
计算1/(2√1 +√2)+1/(3√2+2√3)+……+1/(100√99 +99√100)=?
(1/√2+√1 + 1/√3+√2 +1/√4+√3+.+1/√2010+√2009)*
0-1/(√1+√0)-1/(√2+√1)-1/(√3+√2)+……+1/(√100-√99)
1 √2 √3 √6 1 √2 √3 √6 1 √2 √3 √6 1 √2 √3
(1+√2)分之2+(√2+√3)分之2+(√3+√4)分之2+…+(√99+√100)分之2=?
|1-√2|+|√2-√3|+|√3-√4|+|√4-√5|+...+|√2009-√2010|
1/√3+√2+1/√4+√3+1/√5+√4+1/√6√5