1/(1+√2)+1/(√2+√3)+1/(√3+2)+.+1/(3+√10)=
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1/(1+√2)+1/(√2+√3)+1/(√3+2)+.+1/(3+√10)=
1/(1+√2)+1/(√2+√3)+1/(√3+2)+.+1/(3+√10)=
1/(1+√2)+1/(√2+√3)+1/(√3+2)+.+1/(3+√10)=
分母有理化,=(√2-1)+(√3-√2)+(√4-√3)+...+(√10-√9)=√10-1.
因为1/1+√2=√2-1,1/(√2+√3)=√3-√2。所以原式=√2-1+√3-√2+…+√10-3=√10-1
将每一项都分母有理化得
1/(1+√2)=√2-1
1/(√2+√3)=√3-√2
.......
1/(3+√10)= √10-3
则原式=√2-1+√3-√2+.....+√10-3=√10-1
1/√3+√2+1/√4+√3+1/√5+√4+1/√6√5
|√6-√2|+|1-√2|+|√6-3|
计算:√1/(√1+√2010)+√2/(√2+√2009)+√3/(√3+√2008)+...+√2010/(√2010+√1)
1/(1+√2)+1/(√2+√3)+1/(√3+√4)+.1/(√2012+√2013)
计算:√4-√3/√12+√3-√2/√6+√2-√1/√2=?
1/(√2009+√2008)+1/(√2008+√2007)+.+1/(√3+√2)+1/(√2+1)
√18×√1/2-√3/√12=
(√6-√60)*√3-√1/2=?
化简1/1+√2+1/√2+√3+...+1/√8+√9
1/1+√2+1/√2+√3+…+1/√2013+√2014=
2√2(√2+1/√2)-(√27-√12/√3)
2/3√2-√1/2+√8-√3+√12-√18
1 √2 √3 √6 1 √2 √3 √6 1 √2
2/√3-1
用规律计算:1/1+√2+1/√2+√3+1/√3√4+、、、+1/√2008+√2009+1/√2009√2010+1/√2010+√2011根据(√2-1)*(√2+1)=1(√3-√2)*(√3+√2)=1(√2005-√2004)*(√2005+√2004)=1
1 √32-3√1/2+√22 √12+√27/√3-√1/3×√123 √50+√30/√8-44 √24+√216/√6+55 (√6-2√15)×√3-6√1/26 √2/3-4√216+43√1/67 √8+√30-√28 √1/7+√63-√1129 √40-5√1/10+√1010 √2+√8/√2
√(1-√2)^2+√(√2-√3)^2.+√(√99-√100)√(1-√2)^2+√(√2-√3)^2....+√(√99-√100)^2
|√6|-√2+√(√2-1)² -√(√6-3)²