Bài 2 :

Ta có :

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\left(\frac{1}{16}-\frac{1}{32}\right)\)

    \(=1-\frac{1}{32}\)

và \(B=\frac{2003}{2004}=1-\frac{1}{2004}\)

Vì \(\frac{1}{32}>\frac{1}{2004}\) nên A < B

A<B

mình đang cần gấp

Bài 1:

$M=3.4.5+4.5.6+...+13.14.15$

$4M=3.4.5(6-2)+4.5.6(7-3)+....+13.14.15(16-12)$

$=-2.3.4.5+3.4.5.6-3.4.5.6+4.5.6.7+....-12.13.14.15+13.14.15.16$

$=-2.3.4.5+13.14.15.16=43560$

$M=43560:4=10890$

Bài 2:

a.

$3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}$

$=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{100-97}{97.100}$

$=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}$

$=1-\frac{1}{100}=\frac{99}{100}$

$M=\frac{99}{100}:3=\frac{33}{100}$

giup minh nhe

Chị sẽ giúp em nốt mấy bài này, em còn nhận ra chị ko vậy?

\(A=\frac{2}{1x2}+\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{99x101}\)

\(A=2x\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{99x101}\right)\)

\(A=2x\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(A=2x\left(1-\frac{1}{101}\right)=2x\frac{100}{101}=\frac{200}{101}\)

------------------------------

\(B=\left(1+\frac{1}{2}\right)x\left(1+\frac{1}{3}\right)x\left(1+\frac{1}{4}\right)x...x\left(1+\frac{1}{2016}\right)\)

\(B=\frac{3}{2}x\frac{4}{3}x\frac{5}{4}x...x\frac{2017}{2016}\) (rút gọn từ trên tử xuống dưới mẫu nhé)

\(B=\frac{2017}{2}\)

-------------------------------

\(C=\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+...+\frac{3}{64x67}\)

\(C=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{64}-\frac{1}{67}\)

\(C=1-\frac{1}{67}=\frac{67}{67}-\frac{1}{67}=\frac{66}{67}\)

--------------------------------

\(D=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x...x\left(1-\frac{1}{20}\right)\)

\(D=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{19}{20}\)(chỗ này cũng rút gọn từ trên tử xuống dưới mẫu)

\(D=\frac{1}{20}\)

Đề  phải như này nhé

\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{99.100}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}\right)\)

  \(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\right)\)

   \(=2.\left(1-\frac{1}{100}\right)\)

    \(=2.\frac{99}{100}\)

a) \(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+...+\dfrac{2}{97\times100}\)

\(=2.\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{97\times100}\right)\)

\(=2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(=2.\left(1-\dfrac{1}{100}\right)\)

\(=2.\dfrac{99}{100}\)

\(=\dfrac{99}{50}\)

_____

b) \(\dfrac{3}{1\times5}+\dfrac{3}{5\times9}+\dfrac{3}{9\times13}+...+\dfrac{3}{97\times101}\)

\(=3.\left(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{97\times101}\right)\)

\(=3.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

\(=3.\left(1-\dfrac{1}{101}\right)\)

\(=3.\dfrac{100}{101}\)

\(=\dfrac{300}{101}\)

dấu " . "  là dấu nhân nha bn, nãy viết nhanh quên lớp 

\(=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{31\cdot34}\right)\)

\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)\)

\(=\dfrac{2}{3}\cdot\dfrac{33}{34}=\dfrac{11}{17}\)

\(-\frac{12}{35}\div\frac{7}{11}-\frac{23}{35}\div\frac{7}{11}-\frac{5}{11}\)

\(=\left(-\frac{12}{35}-\frac{23}{35}\right)\div\frac{7}{11}-\frac{5}{11}\)

\(=-1\div\frac{7}{11}-\frac{5}{11}\)

\(=-\frac{11}{7}-\frac{5}{11}\)

\(=-\frac{156}{77}\)

\(2-\frac{11}{9}\div\frac{5}{14}-\frac{7}{9}\times\frac{14}{5}\)

\(=2-\frac{154}{45}-\frac{98}{45}\)

\(=-\frac{64}{45}-\frac{98}{45}\)

\(=-\frac{18}{5}\)

Dấu \(.\)là dấu nhân 

Ta có : 

\(E=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)

\(\Rightarrow E=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{2}{100.103}\right)\)

\(\Rightarrow E=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(\Rightarrow E=\frac{2}{3}.\left(1-\frac{1}{103}\right)\)

\(\Rightarrow E=\frac{2}{3}.\frac{102}{103}\)

\(\Rightarrow E=\frac{68}{103}\)

Vậy \(E=\frac{68}{103}\)

~ Ủng hộ nhé 

\(E=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+...+\frac{2}{100\cdot103}\)

\(E=2\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{100\cdot103}\right)\)

Gọi tổng trong ngoặc là F

\(\Rightarrow3F=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{100\cdot103}\)

\(\Rightarrow3F=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)

\(\Rightarrow3F=1-\frac{1}{103}=\frac{102}{103}\)

\(\Rightarrow F=\frac{102}{103\cdot3}=\frac{34}{103}\)

\(\Leftrightarrow E=2\cdot\frac{34}{103}=\frac{68}{103}\)

Vậy......

E = 2/1×4 + 2/4×7 + 2/7×10 + ... + 2/100×103

E = 2( 1/1×4 + 1/4×7 + 1/7×10 + ... + 1/100×103)

3E = 2( 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/100×103 )

3E = 2(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ...+ 1/100-1/103)

3E = 2( 1 - 1/103 )

3E = 2 × 102/103

3E = 204/103

=> E = 204/103 : 3

     E = 68/103

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