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* Công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\left(\frac{3}{6}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{20.21.22}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{20.21}-\frac{1}{21.22}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{21.22}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{462}\right)\)
\(=\frac{1}{2}.\left(\frac{231}{462}-\frac{1}{462}\right)\)
\(=\frac{1}{2}.\frac{230}{462}\)
\(=\frac{115}{462}\)
Chúc bạn học tốt !!!
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(A=\frac{1}{2}.\frac{370}{741}\)
\(A=\frac{185}{741}\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(\Leftrightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
Tự tính tiếp nha =)) mỏi tay quá
Ta có : \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(\Leftrightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{18.19}-\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{19.20}=\frac{189}{380}\)
\(\Rightarrow B=\frac{189}{760}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{380}\right)\)
\(=\frac{1}{2}.\frac{189}{380}=\frac{189}{760}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{2018\cdot2019\cdot2020}\)
\(=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{2018\cdot2019\cdot2020}\right]\)
\(=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\right]\)
Đến đây tự tính được rồi:v
Đặt tổng trên là A
Ta có:
\(2A=2\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{2018\cdot2019\cdot2020}\right)\)
\(=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{2018\cdot2019\cdot2020}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\)
\(=\frac{1}{2}-\frac{1}{2019\cdot2020}\)
\(A=\left(\frac{1}{2}-\frac{1}{2019\cdot2020}\right)\div2\)
*Làm tiếp*
\(#Louis\)
Ta có công thức:
\(\frac{a}{c.\left[c+1\right].\left[c+2\right]}=\frac{a}{2}\left[\frac{1}{c.\left[c+1\right]}-\frac{1}{\left[c+1\right].\left[c+2\right]}\right]\)
vậy
\(C=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{11.12}-\frac{1}{12.13}\right]\)
\(C=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{12.13}\right]\)
\(C=\frac{1}{2}.\frac{77}{156}=\frac{77}{312}\)
mình làm đầu tiên đó,
Chúc bạn học tốt !
\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{11.12.13}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{11.12}-\frac{1}{12.13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{156}\right)\)
\(=\frac{1}{2}\cdot\frac{77}{156}\)
\(=\frac{77}{312}\)
Mình không chép đề bài nhé :
Gọi biểu thức là A :
Ta có : 2A=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\)
= \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\)
=\(\frac{1}{1.2}-\frac{1}{49.50}\)( Rút gọn đi ta được cái này )
=1/2 - 1/2450
Vậy A = (1/2 - 1/2450):2
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{38.39}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{1482}\right)\)
\(=\frac{1}{2}.\left(\frac{741}{1482}-\frac{1}{1482}\right)\)
\(=\frac{1}{2}.\frac{740}{1482}\)
\(=\frac{185}{741}\)
Chúc bạn học tốt !!!
Đặt 1/1.2.3 + 1/2.3.4 + ...+ 1/37.38.39 = A
Ta có : 2A = 2/1.2.3 + 2/2.3.4 +...+ 2/37.38.39
2A = 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ...+ 1/37.38 - 1/38.39
2A = 1/1.2 - 1/38.39
2A = 740/1482 = 370/741
A= 370/741 . 1/2 =........
\(\frac{3x}{5}=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{6.7.8}\)
Ta có: \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{6.7.8}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{6.7.8}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{6.7}-\frac{1}{7.8}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{7.8}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{56}\right)\)
\(=\frac{1}{2}.\frac{27}{56}=\frac{27}{112}\)
\(\frac{3x}{5}=\frac{27}{112}\)
\(\Rightarrow3x=\frac{27.5}{112}\)
\(\Rightarrow3x=\frac{135}{112}\)
\(\Rightarrow x=\frac{45}{112}\)
~Học tốt~
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{20\cdot21\cdot22}=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{20\cdot21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{20\cdot21}-\frac{1}{21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{21\cdot22}\right)\)
\(=\frac{1}{2}\left(\frac{231}{462}-\frac{1}{462}\right)=\frac{1}{2}\cdot\frac{230}{462}=\frac{1}{2}\cdot\frac{115}{231}=\frac{115}{462}\)