Tính: B= 1x2+2x3+3x4+...+2009x2010
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tinh M=1x2+2x3+3x4+...+2009x2010
1/1x2 + 1/2x3 + 1/3x4 + ............. + 1/2009x2010
=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{2009}\)-\(\frac{1}{2010}\)
=1+(\(\frac{-1}{2}\)+\(\frac{1}{2}\))+(\(\frac{-1}{3}\)+\(\frac{1}{3}\))+...+(\(\frac{-1}{2009}\)+\(\frac{1}{2009}\))-\(\frac{1}{2010}\)
=1+0+0+...+0-\(\frac{1}{2010}\)
=1-\(\frac{1}{2010}\)
=\(\frac{2010}{2010}\)-\(\frac{1}{2010}\)
=\(\frac{2009}{2010}\)
lớp 4 ghê nhỉ đã học bài này rùi tui lớp 6 mà mới học bài này
Tính tổng S biết:
S = \(\frac{1}{1x2}\)+ \(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ ... + \(\frac{1}{2008x2009}\)+ \(\frac{1}{2009x2010}\)
S=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2009}\)-\(\frac{1}{2010}\)
S=1-\(\frac{1}{2010}\)
S=\(\frac{2009}{2010}\)
k nha bn
\(S=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2008\times2009}+\frac{1}{2009\times2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}\)
\(=\frac{2009}{2010}\)
Vậy \(S=\frac{2009}{2010}\)
Học tốt #
s = 1- 1/2+ 1/2- 1/3+ 1/3- 1/4 ......-1/2008- 1/2009+ 1/2009- 1/2010
s =1- 1/2010
s = 2009/2010
Tính :B=2/1x2+2/2x3+2/3x4……+2/99x100
B = \(\dfrac{2}{1\times2}\) + \(\dfrac{2}{2\times3}\)+ \(\dfrac{2}{3\times4}\)+...+ \(\dfrac{2}{99\times100}\)
B = 2 \(\times\) ( \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\)+ \(\dfrac{1}{3\times4}\)+....+ \(\dfrac{1}{99\times100}\))
B = 2 \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+...+ \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\))
B = 2 \(\times\) ( \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\))
B = 2 \(\times\) \(\dfrac{99}{100}\)
B = \(\dfrac{99}{50}\)
1/1x2+1/2x3+1/3x4+1/24x25
1/1x2+ 1/2x3+1/3x4+1/24x25
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=1-\dfrac{1}{25}\)
\(=\dfrac{24}{25}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+.....+\frac{1}{2009x2010}+\frac{1}{2010x2011}\)
Mấy bạn giúp mình nhé , mình đang gấp , 9h mình cần rồi , nhớ giải chi tiết nhé , thanks nhiều ( sẽ hậu tạ )
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}\)
\(=\frac{2010}{2011}\)
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011
= 1 - 1/2011
= 2010/ 2011
Đáp số: 2010/2011
Chúy ý công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2010 - 1/2011
= 1 - 1/2011
= 2010/2011
Đáp sô: 2010/2011
Chú ý công thưc: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
tính nhanh giá trị biểu thức:a) 1/1x3 + 1/3x5 + 1/5x7 +...+ 1/2007x2009
b) 1/1x2 + 1/2x3 + 1/3x4 +...+ 1/2009x2010
\(A=\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{2007.2009}\)
\(2.A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2007.2009}\)
\(2.A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2007}-\frac{1}{2009}\)
\(2.A=1-\frac{1}{2009}=\frac{2008}{2009}\)
A=1004/2009
Tính tổng: 1x2+2x3+3x4+...+98x99
A= 1x2+2x3+3x4+...+98x99 A x 3= 1x2 x (3-0) +2x3x (4-1)+3x4 x (5-2)+...+98x99x (100-97) = 1x2x3+2x3x4+......98x99x100- (1x2x0+ 2x3x1+....+ 98x99x97) = 98x99x100
Tính tổng: 1x2+2x3+3x4+...+98x99
A= 1x2+2x3+3x4+...+98x99
A x 3= 1x2 x (3-0) +2x3x (4-1)+3x4 x (5-2)+...+98x99x (100-97)
= 1x2x3+2x3x4+......98x99x100- (1x2x0+ 2x3x1+....+ 98x99x97)
= 98x99x100.