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A=.....
=\(7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+.....+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
MẤY PHẦN SAU CX TÁCH MẪU RA RÙI LÀM NHƯ VẬY
TỰ LÀM NHE
\(B=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+...+\frac{1}{30\cdot33}\)
\(B=\frac{1}{3}\cdot\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+...+\frac{3}{30\cdot33}\right)\)
\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(B=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)
\(C=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(C=\left(1-\frac{1}{1\cdot2}\right)+\left(1-\frac{1}{2\cdot3}\right)+...+\left(1-\frac{1}{9\cdot10}\right)\)
\(C=9-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)\)
\(C=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(C=9-\left(1-\frac{1}{10}\right)\)
\(C=9-\frac{9}{10}=\frac{81}{10}\)
1) A=7/10.11+7/11.12+7/12.13+...+7/69.70
A=7.(1/10.11+1/11.12+1/12.13+...+1/69.70)
A= 7.(1/10-1/11+1/11-1/12+1/12-1/13+...+1/69-1/70)
A= 7.(1/10-1/70)
A=7.3/35=3/5
2)B=1/25.27+1/27.29+1/29.31+...+1/73.1/75
B=1/25-1/27+1/27-1/29+1/29-1/31+...+1/73-1/75
B=1/25-1/75=2/75
A = 7/ 10.11 + 7/ 11.12 + 7/ 12.13 + .... + 7/69.70
(1/7).A=1/10.11+1/11.12+...+1/69.70
=1/10-1/11+1/11-1/12+...+1/69-1/70
=1/10-1/70=3/35
=>A=7.(3/35)
=3/5
2 ) B = 1/ 25.27 + 1/ 27.29 + 1/29.31+ ......+ 1/ 73.75
=>(1/2).B=2/25.27+...+2.73.75
=1/25-1/27+...+1/73-1/75
=1/25-1/75
=2/75
=>B=4/75
\(A=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=\dfrac{7.60}{700}=\dfrac{420}{700}=\dfrac{3}{5}\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{25}-\dfrac{1}{75}\right)=\dfrac{1}{75}\)
\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{120}\)
\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)
\(A=\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{15.16}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)
\(A=2.\dfrac{3}{16}\)
\(A=\dfrac{3}{8}\)
\(B=\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+...+\dfrac{4}{107.111}\)
\(B=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{107}-\dfrac{1}{111}\)
\(B=\dfrac{1}{3}-\dfrac{1}{111}\)
\(B=\dfrac{12}{37}\)
A = \(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
=\(7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{70}\right)\)
=\(7.\frac{3}{35}\)
=\(\frac{3}{5}\)
B=\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
=\(\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)
=\(\frac{1}{2}.\frac{2}{75}\)
=\(\frac{1}{75}\)
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a) 17.13+17.42-17.35
=17.(13+42-35)
=17.20=340
b) [25.(18-42)-10]:4+6
=(25.2-10):4+6
=40:4+6=16
c) 36:32+23.22-32.3
=34+25-33
=81+32-27=86
d) B=3.42-22.3
=3.(16-4)
=3.12=36
e)20220+3.[52.10-(23-13)2]
=1+3.(250-100)
=1+450=451
g) 27.77+24.27-27
=27.(77+24-1)
=27.100=2700
h) 5.23+79:77-12020
=40+72-1
=89-1=88
i) 120:{54[50:2+(32-2.4)]}
=120:[54(25+1)]
=120:1404=10/117
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2023\cdot2024}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)
\(=1-\dfrac{1}{2024}=\dfrac{2023}{2024}\)
\(B=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{85\cdot89}\)
\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{85}-\dfrac{1}{89}\)
\(=1-\dfrac{1}{89}=\dfrac{88}{89}\)
\(C=\dfrac{7}{10\cdot11}+\dfrac{7}{11\cdot12}+...+\dfrac{7}{69\cdot70}\)
\(=7\left(\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+...+\dfrac{1}{69\cdot70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7\cdot\dfrac{6}{70}=\dfrac{42}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)
\(D=\dfrac{1}{18}+\dfrac{1}{54}+...+\dfrac{1}{990}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+...+\dfrac{3}{30\cdot33}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2023\cdot2024}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)
\(=1-\dfrac{1}{2024}\)
\(=\dfrac{2023}{2024}\)
\(B=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{85\cdot89}\)
\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{85}-\dfrac{1}{89}\)
\(=1-\dfrac{1}{89}\)
\(=\dfrac{88}{89}\)
\(C=\dfrac{7}{10\cdot11}+\dfrac{7}{11\cdot12}+...+\dfrac{7}{69\cdot70}\)
\(=7\left(\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+...+\dfrac{1}{69\cdot70}\right)\)
\(=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{70}\right)\)
\(=7\cdot\dfrac{6}{70}\)
\(=\dfrac{3}{5}\)
\(D=\dfrac{1}{18}+\dfrac{1}{54}+...+\dfrac{1}{990}\)
\(=\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+\dfrac{1}{9\cdot12}+...+\dfrac{1}{30\cdot33}\)
\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+...+\dfrac{3}{30\cdot33}\right)\)
\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{6}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}\cdot\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)