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= \(\left(1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{38}-\frac{1}{38}+\frac{1}{39}\right)\)
= 1 + \(1+\frac{1}{39}=\frac{40}{39}\)
chỗ " 1 + " phía trước là bỏ
ngay chỗ dấu bằng thứ hai
=1-1/2+1/2-1/3+1/3-1/4+1*4-1/5+1/5+1/6=1-1/6=5/6
ok xong
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}=1-\frac{1}{2}+\)+.... \(+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}=\frac{5}{6}\)
Tính nhanh :
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=2.\left(\frac{1}{1.2}+\frac{2}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\left(\frac{20}{20}-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}\)
\(=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}\)
\(=\frac{19}{10}\)
C=(1+1+1+...+1)+(1/1*3+1/2*4+1/3*5+...+1/2015*2017+1/2015*2017)
C=2015+(2/1*3+2/2*4+2/3*5+...+2/2015*2017+2/2015*2017):2
C=2015+(1-1/3+1/2-1/4+...+1/2015-1/2017+1/2015-1/2017):2
C=2015+(1+1/2-1/2016-1/2017+1/2015-1/2017)
cai nay thi ban tu tinh lay
nho k cho minh voi nhe
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)
\(A=\frac{1}{4.6}+\frac{1}{10.12}+\frac{1}{18.20}+...+\frac{1}{810.812}\)
.......
~ Chúc học tốt ~
Ai ngang qua xin để lại 1 L - I - K - E
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{27.28.29.30}\)
\(3A=3.\left(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+......+\frac{1}{27.28.29.30}\right)\)
\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+..........+\frac{3}{27.28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{6}-\frac{1}{24360}\)
\(3A=\frac{1353}{8120}\)
\(A=\frac{1353}{8120}:3\)
\(A=\frac{451}{8120}\)
\(B1\)
\(=\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{37}-\frac{1}{38}-\frac{1}{39}\)
\(=1-\frac{1}{39}\)
\(=\frac{38}{39}\)
\(B2\)
\(=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+.....+\frac{1}{99\cdot100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+......+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{25}{100}-\frac{1}{100}\)
\(=\frac{24}{100}\)
\(=\frac{6}{25}\)
Bài 1 :
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}\)
\(=\frac{370}{741}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{14.15.16}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{14.15.16}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{14.15}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{240}\right)\)
\(=\frac{1}{2}.\frac{119}{240}\)
\(=\frac{119}{480}\)
Bài làm:
Ta có:\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{14.15.16}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{14.15.16}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{14.15}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{15.16}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{240}\right)\)
\(=\frac{1}{2}.\frac{119}{240}=\frac{119}{480}\)
A = 6.(1/1.3+1/3.5+...+1/2015.2017)
= 6.(1/1-1/3+1/3-1/5+...+1/2015-1/2017)
= 6.(1/1-1/2017)
= 6.2016/2017
=12096/2017
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{2016\cdot2017}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}=\frac{2016}{2017}\)
= 1/1-1/2+1/2-1/3+1/3-............-1/2017
=1-1/2017
=2016/2017