- Tính A = 10/56 + 10/140 + 10/260 + ... +10/1400
- Cho A = (1/22 - 1).(1/32 - 1).(1/42 - 1)...(1/1002- 1)
- Cho B = 1/1.2.3.4 + 1/2.3.4.5 + 1/3.4.5.6 + 1/4.5.6.7 + ... + 1/26.27.28.29
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5P=(5-0).1.2.3.4+(6-1).2.3.4.5+...+(101-96).97.98.99.100
5P=1.2.3.4.5-0+2.3.4.5.6-1.2.3.4.5+....+97.98.99.100.101-96.97.98.99.100
5P=97.98.99.100.101
5P=9505049400
S=1901009880
P = 1.2.3.4 + 2.3.4.5 + 3.4.5.6 + 4.5.6.7 + .. + 97.98.99.100
4P = ( 1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + .. + 98.99.100) 4
4P = 1.2.3.(4-0) + 2.3.4(5-1) + 3.4.5(6-2) + 4.5.6(7-3) + 98.99.100(101-97)
4P = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + 4.5.6.7 - 3.4.5.6 + .. 98.99.100.101 - 97.98.99.100
4P = 98.99.100.101
4P= 98.99.100.101/4
Nếu thấy đúng thì tích mk nha
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)
=> \(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+\frac{3}{3.4.5.6}+...+\frac{3}{27.28.29.30}\)
=> \(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+...+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
=> \(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}=\frac{14.29.10-1}{28.29.30}=\frac{4059}{28.29.30}\)
=> \(A=\frac{4059}{28.29.30}:3=\frac{1353}{28.29.30}=\frac{451}{28.29.10}\)
=> \(A=\frac{451}{8120}\)
Các bạn giải giùm mình nhanh nhanh được k? Mình đang cần gấp
\(A=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(A=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\\ A=\frac{1}{75}\)
\(B=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146+150}=\frac{1}{4}\left(\frac{15}{90}-\frac{15}{94}+\frac{15}{94}-\frac{15}{98}+...+\frac{15}{146}-\frac{15}{150}\right)\)
\(B=\frac{1}{4}\left(\frac{15}{90}-\frac{15}{150}\right)=\frac{1}{60}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(A=\frac{1-\frac{1}{3^{100}}}{2}\)
\(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)
\(3B=\frac{5.3}{4.7}+\frac{5.3}{7.10}+\frac{5.3}{10.13}+...+\frac{5.3}{25.28}\)
\(3B=5\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)
\(3B=5\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(3B=5\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(3B=5\cdot\frac{3}{14}=\frac{15}{14}\)
\(B=\frac{15}{14}:3=\frac{5}{14}\)
a) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{100}}}{2}\)
b) \(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(B=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{25.28}\)
\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{5}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+\frac{5}{3}.\left(\frac{1}{10}-\frac{1}{13}\right)+...+\frac{5}{3}.\left(\frac{1}{25}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(B=\frac{5}{3}.\frac{3}{14}\)
\(\Rightarrow B=\frac{5}{14}\)
a) F = \(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
F = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}\right)+\frac{1}{2}.\left(\frac{1}{27}-\frac{1}{29}\right)+\frac{1}{2}.\left(\frac{1}{29}-\frac{1}{31}\right)+...+\frac{1}{2}.\left(\frac{1}{73}-\frac{1}{75}\right)\)
F = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
F = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)
F = \(\frac{1}{2}.\frac{2}{75}\)
F = \(\frac{1}{75}\)
b) G = \(\frac{15}{90.94}+\frac{15}{94.98}+\frac{15}{98.102}+...+\frac{15}{146.150}\)
G = \(\frac{15}{4}.\frac{4}{90.94}+\frac{15}{4}.\frac{4}{94.98}+\frac{15}{4}.\frac{4}{98.102}+...+\frac{15}{4}.\frac{4}{146.150}\)
G = \(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}\right)+\frac{15}{4}.\left(\frac{1}{94}-\frac{1}{98}\right)+\frac{15}{4}.\left(\frac{1}{98}-\frac{1}{102}\right)+...+\frac{15}{4}.\left(\frac{1}{146}-\frac{1}{150}\right)\)
G = \(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+\frac{1}{98}-\frac{1}{102}+...+\frac{1}{146}-\frac{1}{150}\right)\)
G = \(\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\)
G = \(\frac{15}{4}.\frac{1}{225}\)
G = \(\frac{1}{60}\)
<br class="Apple-interchange-newline"><div id="inner-editor"></div>12.4 +14.6 +...+198.100
=12 (22.4 +24.6 +...+298.100 )
<br class="Apple-interchange-newline"><div id="inner-editor"></div>=12 (12 −14 +14 −16 +...+198 −1100 )
<br class="Apple-interchange-newline"><div id="inner-editor"></div>=12 (12 −14 +14 −16 +...+198 −1100 )
<br class="Apple-interchange-newline"><div id="inner-editor"></div>=12 (12 −1100 )=12 .49100 =49200
1056 +10140 +10260 +...+101400 =53 (
Tk mình đi mọi người mình bị âm nè!
ai tk mình mình tk lại cho!!!
1.
A= 5/28 + 5/70 +.....+10/700 = 5/(4.7)+5/(7.10)+....5/(25.28)
3A= 5( 1/4 - 1/7 +1/7-1/10+......+1/25-1/28)
3A= 5 (1/4-1/28)
3A=15/14
A= 5/14
#)Giải :
1. \(A=\frac{10}{54}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
\(A=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)
\(\Rightarrow\frac{3A}{5}=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\)
\(\Rightarrow\frac{3A}{5}=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\)
\(\Rightarrow\frac{3A}{5}=\frac{1}{4}-\frac{1}{28}=\frac{3}{14}\)
\(\Rightarrow A=\frac{3}{14}\times\frac{5}{3}\)
\(\Rightarrow A=\frac{5}{14}\)