\(\frac{11}{1.4}+\frac{11}{4.7}+...+\frac{11}{100.103}\)

\(=\frac{11}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(=\frac{11}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(=\frac{11}{3}\left(1-\frac{1}{103}\right)\)

Tự tính

\(\frac{11}{1.4}+\frac{11}{4.7}+...+\frac{11}{100.103}\)

= \(\frac{11}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

= \(\frac{11}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

= \(\frac{11}{3}.\left(1-\frac{1}{103}\right)\)

= \(\frac{11}{3}.\frac{102}{103}\)

= \(\frac{374}{103}\)

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

= \(\frac{1}{3}\left(1-\frac{1}{x+1}\right)\)

=\(\frac{1}{3}x\left(\frac{x+1}{x+1}-\frac{1}{x+1}\right)\)

= \(\frac{1}{3}x\frac{x}{x+1}\)

dè bài la tim x

Tong cua phep. Tinh la 2009/6030

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{34.37}+\dfrac{1}{37.40}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}\)

\(=1-\dfrac{1}{40}\)

\(=\dfrac{39}{40}\)

1/(1×4) + 1/(4×7) + ... + 1/(34×37) + 1/(37×40)

= 1/3 × (1 - 1/4 + 1/4 - 1/7 + ... + 1/34 - 1/37 + 1/37 - 1/40)

= 1/3 × (1 - 1/40)

= 1/3 × 39/40

= 13/40

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{34.37}+\dfrac{1}{37.40}\)

\(=1-\left(\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}\right)\)

\(=1-\dfrac{1}{40}\)

\(=\dfrac{39}{40}\)

Vậy giá trị cần tìm là: \(\dfrac{39}{40}\)

\(3xA=\dfrac{4-1}{1x4}+\dfrac{7-4}{4x7}+...+\dfrac{37-34}{34x37}+\dfrac{40-37}{37x40}=\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{17}{ }+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}=\)

\(=1-\dfrac{1}{40}=\dfrac{39}{40}\Rightarrow A=\dfrac{39}{40}:3=\dfrac{13}{40}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{34.37}+\dfrac{3}{37.40}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}\right)\)

\(=\dfrac{1}{3}.\dfrac{39}{40}\)

\(=\dfrac{13}{40}\)

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 

Ta có    1/1x4+1/4x7+...+1/2002x2005

  <=>   =1/3.3(1/1x4+1/4x7+...+1/2002x2005)

            =1/3(3/1x4+3/4x7+...+3/2002x2005)

            =1/3(1-1/4+1/4-1/7+...+1/2002-1/2005)

            =1/3(1-1/2005)

            =1/3.2004/2005

            =1.2004/3.2005

            =668/2005

\(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+...+\(\frac{1}{2002.2005}\)=3(\(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+...+ \(\frac{1}{2002.2005}\)):3=(\(\frac{3}{1.4}\)+ \(\frac{3}{4.7}\)+...+ \(\frac{3}{2002.2005}\)):3= (1-\(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)):3=(1-\(\frac{1}{2005}\)) : 3 = \(\frac{668}{2005}\)

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.

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

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

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

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

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