化简 1) √[(1998*1999*2000*2001+1)/4] 2) 1/(2√1+1√2)+1/(3√2+2√3
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/22 04:37:06
化简 1) √[(1998*1999*2000*2001+1)/4] 2) 1/(2√1+1√2)+1/(3√2+2√3)+……+1/(100√99+99√100)
化简 1) √[(1998*1999*2000*2001+1)/4]
2) 1/(2√1+1√2)+1/(3√2+2√3)+……+1/(100√99+99√100)
化简 1) √[(1998*1999*2000*2001+1)/4]
2) 1/(2√1+1√2)+1/(3√2+2√3)+……+1/(100√99+99√100)
1、
令a=1998
则1998*1999*2000*2001+1
=a(a+1)(a+2)(a+3)+1
=[a(a+3)][(a+1)(a+2)]+1
=(a²+3a)[(a²+3a)+2]+1
=(a²+3a)²+2(a²+3a)+1
=(a²+3a+1)²
所以原式=√[(a²+3a+1)²/4]
=(a²+3a+1)/2
=(1998²+3×1998+1)/2
=3997999/2
2、
1/[(n+1)√n-n√(n+1)]
=[(n+1)√n-n√(n+1)]/[(n+1)√n-n√(n+1)][(n+1)√n+n√(n+1)]
=[(n+1)√n-n√(n+1)]/[n(n+1)²-n²(n+1)]
=[(n+1)√n-n√(n+1)]/n(n+1)
=(n+1)√n/n(n+1)-n√(n+1)/n(n+1)
=√n/n-√(n+1)/(n+1)
所以原式=√1/1-√2/2+√2/2-√3/3+……+√99/99-√100/100
=1-10/100
=9/10
令a=1998
则1998*1999*2000*2001+1
=a(a+1)(a+2)(a+3)+1
=[a(a+3)][(a+1)(a+2)]+1
=(a²+3a)[(a²+3a)+2]+1
=(a²+3a)²+2(a²+3a)+1
=(a²+3a+1)²
所以原式=√[(a²+3a+1)²/4]
=(a²+3a+1)/2
=(1998²+3×1998+1)/2
=3997999/2
2、
1/[(n+1)√n-n√(n+1)]
=[(n+1)√n-n√(n+1)]/[(n+1)√n-n√(n+1)][(n+1)√n+n√(n+1)]
=[(n+1)√n-n√(n+1)]/[n(n+1)²-n²(n+1)]
=[(n+1)√n-n√(n+1)]/n(n+1)
=(n+1)√n/n(n+1)-n√(n+1)/n(n+1)
=√n/n-√(n+1)/(n+1)
所以原式=√1/1-√2/2+√2/2-√3/3+……+√99/99-√100/100
=1-10/100
=9/10
化简 1) √[(1998*1999*2000*2001+1)/4] 2) 1/(2√1+1√2)+1/(3√2+2√3
计算:(√3+1)2001-2(√3+1)2000-2(√3+1)1999+2001=
(1/√2+√1 + 1/√3+√2 +1/√4+√3+.+1/√2010+√2009)*
化简1/1+√2+1/√2+√3+...+1/√8+√9
化简[1/(1+√2)]+[1/(√2+√3)]+`````+[1/(√8+√9)]
√(1-√2)^2+√(√2-√3)^2.+√(√99-√100)
计算:√4-√3/√12+√3-√2/√6+√2-√1/√2=?
1/√3+√2+1/√4+√3+1/√5+√4+1/√6√5
|1-√2|+|√2-√3|+|√3-√4|+|√4-√5|+...+|√2009-√2010|
1/(1+√2)+1/(√2+√3)+1/(√3+√4)+.1/(√2012+√2013)
2√1/3化简
化简:2√8+√1(√27+√3)分之√3 - 2