Ta có : B = \(\frac{4}{1.4}+\frac{4}{4.7}+\frac{4}{7.10}+......+\frac{4}{2014.2017}\)

\(=\frac{4}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{2014.2017}\right)\)

\(=\frac{4}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+......+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(=\frac{4}{3}.\left(1-\frac{1}{2017}\right)\)

\(=\frac{4}{3}.\frac{2016}{2017}=\frac{2688}{2017}\)

thank

sai zùi

\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(B=\frac{5}{3}.\left(1-\frac{1}{2017}\right)\)

\(B=\frac{5}{3}.\frac{2016}{2017}=\frac{10080}{6051}\)

\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{2014.2017}\)

\(3M=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{2014.2017}\right)\)

\(3M=5\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(3M=5\left(1-\frac{1}{2017}\right)\)

\(3M=5.\frac{2016}{2017}\)

\(3M=\frac{10080}{2017}\)

\(\Rightarrow M=\frac{3360}{2017}\)

Ta có: 

\(B=\frac{5}{1.4}+\frac{5}{4.7}+.....+\frac{5}{2014.2017}\)

\(\Rightarrow B.\frac{3}{5}=\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{2014.2017}\)

\(=\left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+......+\left(\frac{1}{2014}-\frac{1}{2017}\right)\)

\(=1-\frac{1}{2017}\)

\(=\frac{2016}{2017}\)

\(\Rightarrow B=\frac{2016}{2017}:\frac{3}{5}\)

\(=\frac{3360}{2017}\)

Vậy B \(=\frac{3360}{2017}\)~~~~~

A = \(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+\(\frac{1}{7.10}\)+...+ \(\frac{1}{2014.2017}\)
3A = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{2014.2017}\)
3A = \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{2014}-\frac{1}{2017}\)
3A= 1 - \(\frac{1}{2017}\)
A = \(\frac{1}{3}-\frac{1}{2017.3}\)
A = \(\frac{672}{2017}\)

Ta có \(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2014.2017}\)

\(\Rightarrow A=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{3}.\left(1-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{3}.\frac{2016}{2017}=\frac{672}{2017}\)

Vậy \(A=\frac{672}{2017}\)

~ Học tốt

# Chiyuki Fujito

672/2017

\(\frac{3A}{4}=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{25.28}.\)

\(\frac{3A}{4}=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{28-25}{25.28}\)

\(\frac{3A}{4}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}=1-\frac{1}{28}=\frac{27}{28}\)

\(A=\frac{27.4}{28.3}=\frac{9}{7}\)

Bạn làm đúng rồi.

a) Ta có: \(A=\dfrac{4}{1\cdot4}+\dfrac{4}{4\cdot7}+\dfrac{4}{7\cdot10}+...+\dfrac{4}{31\cdot34}\)

\(=\dfrac{4}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{31\cdot34}\right)\)

\(=\dfrac{4}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)\)

\(=\dfrac{4}{3}\left(1-\dfrac{1}{34}\right)\)

\(=\dfrac{4}{3}\cdot\dfrac{33}{34}=\dfrac{22}{17}\)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

\(2A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(2A=\frac{1}{1}-\frac{1}{100}\)

\(2A=\frac{99}{100}\Rightarrow A=\frac{99}{100}:2\Rightarrow A=\frac{99}{200}\)

Câu B và C làm tương tự.

bạn Nhi làm sai rồi

\(\frac{2}{2\cdot3}\) sao có thể bằng \(\frac{1}{2}-\frac{1}{3}\) được

\(\frac{1}{2\cdot3}\) mới bằng \(\frac{1}{2}-\frac{1}{3}\)

kết quả là : \(\frac{49}{100}\)

Bài 1 :

\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)

Ta thấy :

\(3=2^2-1\)

\(15=4^2-1\)

\(35=6^2-1\)

.....

\(9999=100^2-1\)

\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)

\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)

\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)

nhanh len nhé mik đang cần gấp ai lam trước mik tích cho

Bài 6 :

\(C=1^2+2^2+...+100^2=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}=\dfrac{100.101.201}{6}=338350\)

Bài 9 :

\(S=1^2+2^2+3^2+...+99^2=\dfrac{99.\left(99+1\right)\left(2.99+1\right)}{6}=\dfrac{99.100.199}{6}=328350\)

4/1.4+4/4.7+4/7.10+4/10.13

= 4/3(3/1.4+3/4.7+3/7.10+3/10.13)

=4/3(1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13)

=4/3(1/1-1/13)

=4/3.12/13

=16/13