a) A= 4/3.7 + 4/7.11 + 4/11.15 +...+ 4/107.111
b) B= 2/15 + 2/35 +2/63+...+ 2/399
c) C= 1/10 + 1/15 + 1/21+ ...+ 1/120
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\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)
\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(A=2.\dfrac{3}{16}\)
\(A=\dfrac{3}{8}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(B=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(B=\dfrac{1}{3}-\dfrac{1}{111}\)
\(B=\dfrac{12}{37}\)
Giải:
A=2/3.7+2/7.11+2/11.15+...+2/n.(n+4)
A=1/2.(4/3.7+4/7.11+4/11.15+...+4/n.(n+4)
A=1/2.(1/3-1/7+1/7-1/11+1/11-1/15+...+1/n-1/n+4)
A=1/2.(1/3-1/n+4)
A=1/6-1/2.(n+4)
⇒A<1/6
Chúc bạn học tốt!
Ta có : \(A=\dfrac{2}{3.7}+\dfrac{2}{7.11}+...+\dfrac{2}{n\left(n+4\right)}\)
\(\Rightarrow4A=\dfrac{8}{3.7}+\dfrac{8}{7.11}+...+\dfrac{8}{n\left(n+4\right)}\)
\(\Rightarrow4A=\dfrac{8}{3.7}+\dfrac{8}{7.11}+...+\dfrac{8}{n\left(n+4\right)}\)\(=\dfrac{2}{3}-\dfrac{2}{7}+\dfrac{2}{7}-\dfrac{2}{11}+...+\dfrac{2}{n}-\dfrac{2}{n+4}=\dfrac{2}{3}-\dfrac{2}{n+4}\)
\(\Rightarrow A=\dfrac{1}{6}-\dfrac{1}{2\left(n+4\right)}\)
- Xét hiệu \(A-\dfrac{1}{6}=-\dfrac{1}{2\left(n+4\right)}< 0\)
Vậy A < 1/6
Ta có ; K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)
\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{90}\)
\(=1+\left(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{9.10}\right)\)
\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1+2\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=1+1-\frac{1}{5}\)(nhân phá ngoặc)
\(=2-\frac{1}{5}\)< 2
Vậy K = \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{45}\)< 2
Tính tổng:a)3+3/5+3/25+3/125+3/625
b)M=4/3.7+4/7.11+4/11.15+...+8/95.99
c)N=1/2+1/6+1/12+1/20+...+1/90
Ta có : \(M=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.....+\frac{4}{95.99}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+......+\frac{1}{95}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
\(A=\frac{4^2}{3.7}+\frac{4^2}{7.11}+\frac{4^2}{11.15}+...+\frac{4^2}{107.111}\)
\(A=\) \(4\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{107.111}\right)\)
\(A=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)
\(A=4\left(\frac{1}{3}-\frac{1}{111}\right)\)
\(A=4.\frac{12}{37}\)
\(A=\frac{48}{37}\)
\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{\left(3x-1\right).\left(3x+3\right)}=\frac{3}{10}\)
=> \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{3x-1}-\frac{1}{3x+3}=\frac{3}{10}\)
=> \(\frac{1}{3}-\frac{1}{3x+3}=\frac{3}{10}\)
=> \(\frac{1}{3x+3}=\frac{1}{3}-\frac{3}{10}\)
=> \(\frac{1}{3x+3}=\frac{1}{30}\)
=> 3x + 3 = 30
=> 3.(x + 1) = 30
=> x + 1 = 10
=> x = 9
mk làm phần a thui nhé
a. A = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6
A = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6
A = 1/2 - 1/6
A= 3/6 - 1/6
A = 1/3
\(B=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\)
\(b=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(b=\frac{1}{2}-\frac{1}{14}\)
\(b=\frac{3}{7}\)
\(d=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)
\(d=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}\)
\(d=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
\(d=1-\frac{1}{11}\)
\(d=\frac{10}{11}\)
\(e=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)
\(e=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+\frac{1}{14\cdot17}+\frac{1}{17\cdot20}\)
\(e=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{17\cdot20}\right)\)
\(e=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\right)\)
\(e=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
\(e=\frac{1}{3}\cdot\frac{9}{20}=\frac{3}{20}\)
a)1/2 + 5/6 + 11/12 + 19/20 + 29/30 + 41/42 + 55/56 + 71/72+89/90
=1-1/2+1-1/6+1-1/12+1-1/20+1-1/30+1-1/42+1-1/56+1-1/72+1-1/90
=9 – (1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)
=9 – [1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+1/(5x6)+1/(6x7)+1/(7x8)+1/(8x9)+1/(9x10)]
=9 – ( 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)
=9 – (1 – 1/10) = 9 – 9/10 = 81/10
b)4/3.7 + 4/7.11 + 4/11.15 + 4/15.19 + 4/19.23 + 4/23.27
=4.(4/3.7 + 4/7.11 + ........+ 4/23.27 )
=1.( 1/3.7 + 1/7.11 + ......+ 1/23.27 )
=1.(1/3 - 1/7 + 1/7 - 1/11 +............ + 1/23 - 1/27 )
=1.(1/3 - 1/27 )
=1.(9/27 - 1/27)
=1.8/27
=8/27
c)1/10+1/40+1/88+1/154+1/138+1/340
=1/2.5 + 1/5.8 + 1/11.8 + 1/11.14 + 1/14.17 + 1/17.20
=1/3. (3/2.5 + 3/5.8 + 3/8.11 + 3/11.14 + 3/14.17 + 3/17.20 )
=1/3. ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 + 1/14 - 1/17 + 1/17 -1/20 )
=1/3. ( 1/2 - 1/20 )
=1/3. 9/20
=3/20
P/S: CHÚC HOK TỐT !
Viết vậy đúng đó em
A = 5/(3.7) + 5/(7.11) + 5/(11.15) + ... + 5/(2019.2023)
= 5/4 . [4/(3.7) + 4/(7.11) + 4/(11.15) + ... + 4/(2019.2023)]
= 5/4 . (1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ... + 1/2019 - 1/2023)
= 5/4 . (1/3 - 1/2023)
= 5/4 . 2020/6069
= 2525/6069
a.\(A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)
\(=\frac{1}{3}-\frac{1}{111}=\frac{37}{111}-\frac{1}{111}=\frac{36}{111}=\frac{12}{37}\)
Vậy A=\(\frac{12}{37}\)
b.\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
\(=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(=\frac{1}{3}-\frac{1}{21}=\frac{7}{21}-\frac{1}{21}=\frac{6}{21}=\frac{2}{7}\)
Vậy \(B=\frac{2}{7}\)
c.\(C=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)
\(\Rightarrow C.\frac{1}{2}=\left(\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\right).\frac{1}{2}\)
\(=\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\)
\(=\frac{1}{4}-\frac{1}{16}=\frac{4}{16}-\frac{1}{16}=\frac{3}{16}\)
Vậy \(C=\frac{3}{16}\)
A = \(\frac{4}{3.7}+\frac{4}{7.9}+...+\frac{4}{107.111}\)
A = \(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{107}-\frac{1}{111}\)
A = \(\frac{1}{3}-\frac{1}{111}\)=\(\frac{12}{37}\)
2 câu sau tương tự. Mik ngại làm lắm -_-