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\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{2002^2}+\dfrac{1}{2003^2}< 1\)
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TT
1 tháng 12 2017
Chữa lại đề.Bạn xem lại đề xem đúng chưa nhé!
\(D=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}+\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}+\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}+\dfrac{3}{2004}}\)
\(D=\dfrac{1.\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}\right)}{5.\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}\right)}-\dfrac{2.\left(\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)}{3\left(\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)}\)
\(D=\dfrac{1}{5}-\dfrac{2}{3}\)
\(D=-\dfrac{7}{15}\)
Cái này học lâu rồi.Bạn xem lại xem mình làm đúng chưa nhé!
NN
15 tháng 11 2017
1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
15 tháng 11 2017
b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
20 tháng 3 2023
1:
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
...
\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}\)
=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+..+\dfrac{1}{7\cdot8}\)
=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{7}{8}< 1\)
19 tháng 11 2017
8,A=\(\dfrac{9}{10}-\left(\dfrac{1}{10\times9}+\dfrac{1}{9\times8}+\dfrac{1}{8\times7}+...+\dfrac{1}{2\times1}\right)\)
=\(\dfrac{9}{10}-\left(\dfrac{1}{10}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{8}+...+\dfrac{1}{2}-1\right)\)
=\(\dfrac{9}{10}-\left(\dfrac{1}{10}-1\right)\)
=\(\dfrac{9}{10}-\dfrac{\left(-9\right)}{10}\)
=\(\dfrac{9}{5}\)
31 tháng 10 2022
6:
\(4D=2^2+2^4+...+2^{202}\)
=>3D=2^202-1
hay \(D=\dfrac{2^{202}-1}{3}\)
7: \(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{32}{99}=\dfrac{16}{99}\)
18 tháng 9 2021
Bài 1:
a) \(\left|3x-5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x=-2004\)( do \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\))
Bài 2:
a) \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=1\)
b) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)
\(=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)\)
\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)
18 tháng 9 2021
Bài 1:
a) \(\left|3x-5\right|=4\) (1)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\) \(\left(do\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\) \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)
\(\Leftrightarrow x=-2004\)
=> A<1/1.2 + 1/2.3 + ....+ 1/2001.2002 + 1/2002.2003
=> A< 1- 1/2 + 1/2 - 1/3 + .... + 1/2001 - 1/2002 + 1/2002 - 1/2003
=>A< 1 - 1/2003 < 1
=> A< 1