A = 2/1x2x3 + 2/2x3x4 + 2/3x4x5 + ... + 2/36x37x38 + 2/37x38x39

A = 1/1x2 - 1/2x3 + 1/2x3 - 1/3x4 + 1/3x4 - 1/4x5 + ...+ 1/36x37 - 1/37x38 + 1/37x38 - 1/38x39

A = 1/2 - 1/38x39

A = 370/741

Tớ ko chắc là đúng đâu

bài 

Nguyễn Hà Thương sai chưa chia 2

A=\(\frac{1}{1x2x3}+\frac{1}{2x3x4}+...+\frac{1}{37x38x39}\)

=\(\frac{1}{2}x\left(\frac{1}{1x2}-\frac{1}{2x3}+\frac{1}{2x3}-\frac{1}{3x4}+...+\frac{1}{37x38}-\frac{1}{38x39}\right)=\frac{1}{2}x\left(\frac{1}{2}-\frac{1}{38x39}\right)=\frac{185}{741}\)

\(=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{37\times38}-\frac{1}{38\times39}\)

\(=\frac{1}{1\times2}-\frac{1}{38\times39}=\frac{1}{2}-\frac{1}{1482}=\frac{370}{741}\)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{36.37.37}\)

\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{36.37.38}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{36.37}-\frac{1}{37.38}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{37.38}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{1406}\right)\)

\(=\frac{1}{2}.\frac{351}{703}\)

\(=\frac{351}{1046}\)

\(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....-\frac{1}{36}+\frac{1}{37}-\frac{1}{38}+\frac{1}{38}-\frac{1}{39}\)

=\(\frac{1}{1}-\frac{1}{39}\)

=\(\frac{38}{39}\)

đảm bảo đúng

\(A=\frac{370}{741}\)

A=\(\frac{370}{741}\)

\(2C=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(2C=\frac{1}{1.2}-\frac{1}{38.39}\)
\(C=\frac{617}{1482}\)

\(3D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3D-D=1-\frac{1}{3^8}\)
\(D=\frac{1}{2}-\frac{1}{2.3^8}\)

Ta có:\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{37.38}-\frac{1}{38.39}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{38.39}\right)\)

b,\(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)

\(\Rightarrow3D=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^7}\)

\(\Rightarrow2D=1-\frac{1}{3^8}\)

\(\Rightarrow D=\frac{3^8-1}{3^8}:2\)

Ta có : 

\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{4.5.6}+...+\frac{1}{37.38.39}\)

\(2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\)

\(2C=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)

\(2C=\frac{1}{2}-\frac{1}{38.39}\)

\(C=\frac{\frac{1}{2}-\frac{1}{38.39}}{2}\)

\(C=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{38.39}\right)\)

\(C=\frac{1}{2}.\frac{370}{741}\)

\(C=\frac{185}{741}\)

=1/1.2.3+1/2.3.4+....+1/37.38.39

=1/1.2-1/2.3+1/2.3-1/3.4+...+1/37.38-1/38.39

=1/1.2-1/38.39

=1/2-1/38.39

=370/7410

bạn ơi, mình dùng dấu nhân bằng dấu chấm bạn nha(  còn dấu ... thì bình thường)

mình chắc chắn đúng. cái này bạn học toán phần nâng cao lớp 5 dạng sai phân.

chúc bạn học giỏi

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+......+\frac{1}{48.49.50}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{49.50}\right)\)

\(=\frac{1}{2}.\frac{612}{1225}=\frac{612}{2450}=\frac{306}{1225}\)

Do not ask why hay quá!

Đặt \(T=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)

Ta xét:

\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{1}{1.2.3}\);\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{1}{2.3.4}\);. . . ; \(\frac{1}{48.49}-\frac{1}{49.50}=\frac{1}{48.49.50}\)

 Rút ra dạng tổng quát,ta có: (mình nói thêm nhé)

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

\(\Rightarrow2T=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\)

Ta nhận thấy: \(-\frac{1}{2.3}+\frac{1}{2.3}=0\);\(-\frac{1}{3.4}+\frac{1}{3.4}=0\);.....

\(\Rightarrow2T=\frac{1}{1.2}-\frac{1}{49.50}=\frac{612}{1225}\)

\(\Rightarrow T=\frac{612}{\frac{1225}{2}}=\frac{306}{1225}\)

Vậy .. . . 

ta co:1/1*2*3=(1/1*2-1/2*3):2
1/2*3*4=(1/1*2-1/2*3):2
...
cu nhu the cho den:
1/98*99*100=(1/98*99-1/99*100):2
suy ra : 1/1*2*3+1/2*3*4+1/3*4*5+...+1/98*99*100
=(1/1*2-1/2*3):2+(1/2*3-1/3*4):2+...+(1/98*99-1/99*100):2
=(1/1*2-1/2*3+1/2*3-1/3*4+...+1/98*99-1/99*100):2
=(1/1*2-1/99*100):2
=(1/2-1/9900)
=(4950/9000-1/9000):2
=4949/9000:2
=4949/18000
học tốt

cái đuôi ak ko hiểu còn cái đầu thì dễ 

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.......+\frac{1}{8.9.10}\)

\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+......+\frac{2}{8.9.10}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+.......+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)