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b. Gọi tổng trên là A
=> A=10,11+11,12+12,13+....+97,98+98,99+99,100
=> A-99,1=10,11+11,12+12,13+.....+97,98+98,99
SSH của A-99,1 là
(98,99-10,11):1,01+1=89(SH)
Giá trị của A-99,1 là
\(\frac{89}{2}.\left(10,11+98,99\right)=4854,95\)
Vì A-99.1=4854,95
=> A=4854,95+99,1
=>A=4954.05
****
a) \(\frac{2011.2010-1}{2009.2011+2010}=\frac{2011.2009+2011-1}{2009.2011+2010}=\frac{2011.2009+2010}{2009.2011+2010}=1\)
. là nhân nha
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A = \(\dfrac{1}{3\times5}\) + \(\dfrac{1}{5\times7}\) + \(\dfrac{1}{7\times9}\)+...+ \(\dfrac{1}{2009\times2011}\)
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\)+ \(\dfrac{2}{7\times9}\)+...+ \(\dfrac{1}{2009\times2011}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+ \(\dfrac{1}{2009}\) - \(\dfrac{1}{2011}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{2011}\))
A = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2008}{6033}\)
A = \(\dfrac{1004}{6033}\)
\(\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{2}{7\times9}+..+\dfrac{1}{2009\times2011}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\\ =\dfrac{1}{3}-\dfrac{1}{2011}\)
Đến đây bn tự tính nhé.
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Lời giải:
$t=\frac{2009\times 2010+2000}{2011\times 2010-2020}$
$=\frac{2010\times (2011-2)+2000}{2011\times 2010-2020}$
$=\frac{2010\times 2011-2\times 2010+2000}{2011\times 2010-2020}$
$=\frac{2010\times 2011-2020}{2011\times 2010-2020}=1$
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Ta có:
T = \(\frac{2009\times2010+2000}{2011\times2010-2020}=\frac{2009\times2010+2010-10}{2010\times2010+2010-2020}=\frac{2010\times2010-10}{2010\times2010-10}=1\)
Vậy T = 1
2009×2010+2000/2011×2010-2020
=2009×2010+2000/2009×2010+2×2010-2020
=2009×2010+2000/2009×2010+4020-2020
=2000/4020-2020
=2000/2000
=1
K đúng cho mk nha !!!
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\(\frac{1}{5.7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{2009\cdot2011}+\frac{1}{x}=\frac{1}{5}\cdot0,5\)
\(=\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+...+\frac{2011-2009}{2009\cdot2011}+\frac{1}{x}=\frac{1}{10}\)
\(=\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2009}-\frac{1}{2011}\right)\right]+\frac{1}{x}=\frac{1}{10}\)
\(=\left[\frac{1}{2}\cdot\left(\frac{1}{5}-\frac{1}{2011}\right)\right]+\frac{1}{x}=\frac{1}{10}\)
\(=\left(\frac{1}{2}\cdot\frac{2006}{10055}\right)+\frac{1}{x}=\frac{1}{10}\)
\(=\frac{1003}{10055}+\frac{1}{x}=\frac{1}{10}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{10}-\frac{1003}{10055}\)
\(\frac{1}{x}=\frac{1}{4022}\)
\(\Rightarrow x=1\div\frac{1}{4022}=4022\)
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Ta thấy : 1 x 3 x 5 x 7 x 9 có tận cùng là 5
Các số sau nhân cũng có tận cùng là 5 nên chữ số tận cùng của tích 1 x 3 x 5 x 7 x 9 x...x 2011 cũng là 5
Đáp số : tận cùng là 5
\(\frac{2009.2011-1}{2011.2010-2012}=\frac{2009.2011-1}{2011.\left(2009+1\right)-2012}\)
\(=\frac{2009.2011-1}{2011.2009+2011-2012}\)
\(=\frac{2009.2011-1}{2009.2011-1}\)
\(=1\)
\(\frac{2009.2011-1}{2011.2010-2012}=\frac{2009.2011-1}{2011.2009-2011-2012}=\frac{2009.2011-1}{2011.2009-1}=1\)