\(D=1.2.3+2.3.4+...+8.9.10\)

\(4D=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)

\(4D=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)

\(4D=8.9.10.11\)

\(D=\frac{8.9.10.11}{4}=\frac{7920}{4}=1980\)

Đặt A=1.2.3+2.3.4+...+8.9.10

\(4A=1.2.3.4+2.3.4.4+...+8.9.10.4\)

\(4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)

\(4A=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)

\(4A=\left(1.2.3.4+2.3.4.5+...+8.9.10.11\right)-\left(0.1.2.3+1.2.3.4+...+7.8.9.10\right)\)

\(4A=8.9.10.11-0.1.2.3\)

\(4A=8.9.10.11\)

\(A=2.9.10.11\)

\(\Rightarrow A=1980\)

cảm ơn ST

\(B=\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+...+\dfrac{1}{8.9.10}\)

\(B=2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(B=2.\left(1-\dfrac{1}{10}\right)\)

\(B=2.\dfrac{9}{10}\)

\(B=\dfrac{9}{5}\)

anh ơi , đại học rồi mà ko giải đc bài này ạ?

hmm

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}...+\frac{2}{8.9.10}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{11}{45}\)

     \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)\)

\(=\frac{1}{2}.\frac{22}{45}\)

\(=\frac{11}{45}\)

<=>4S=1.2.3.4 + 2.3.4.4+3.4.5.4+.....+8.9.10.4
<=>4S =1.2.3.4 + 2.3.4.(5-1) + 3.4.5.(6-2)+......+8.9.10.(11-7)
<=>4S=1.2.3.4 + 2.3.4.5 -1.2.3.4+3.4.5.6- 2.3.4.5+......+8.9.10.11 - 7.8.9.10
<=> 4S=8.9.10.11
<=>S=1980

Làm giúp mình đi

MONG CAC BAN GIUP ĐỠ

GIUP MÌNH VÓI

\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right).x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{8.9.10}\right).x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-...-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.\dfrac{22}{45}.x=\dfrac{22}{45}\)

=> \(\dfrac{1}{2}.x=1\)

=> \(x=2\)

Vậy x = 2

Chúc bạn học tốt !!!