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DP
5 tháng 7 2017
\( A=\frac{3^3}{6\cdot11}+\frac{3^3}{11\cdot16}+\frac{3^3}{16\cdot21}+....+\frac{3^3}{91\cdot96}\)
\(A=\frac{3^3}{5}\cdot\left(\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{91}-\frac{1}{96}\right)\)
\(A=\frac{27}{5}\cdot\left(\frac{1}{6}-\frac{1}{96}\right)\)
\(A=\frac{27}{5}\cdot\frac{5}{32}=\frac{27}{32}\)
11 tháng 3 2020
\(D=\frac{3}{11\cdot16}+\frac{3}{16\cdot21}+\frac{3}{21\cdot26}+....+\frac{3}{61\cdot66}\)
\(\frac{5}{3}D=\frac{5}{3}\left(\frac{3}{11\cdot16}+\frac{3}{16\cdot21}+\frac{3}{21\cdot26}+.....+\frac{3}{61\cdot66}\right)\)
\(\frac{5}{3}D=\frac{5}{11\cdot16}+\frac{5}{16\cdot21}+\frac{5}{21\cdot26}+....+\frac{5}{61\cdot66}\)
\(\frac{5}{3}D=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+....+\frac{1}{61}-\frac{1}{66}\)
\(\frac{5}{3}D=\frac{1}{11}-\frac{1}{66}\)
\(\frac{5}{3}D=\frac{5}{66}\)
\(D=\frac{5}{66}:\frac{5}{3}=\frac{5}{66}\cdot\frac{6}{5}=\frac{1}{11}\)
E
11 tháng 3 2020
D = 3/11.16 + 3/16.21 + 3/21.26 + ...... + 3/61.66
D = \(\frac{3}{5}\) . ( \(\frac{5}{11.16}\)+ \(\frac{5}{16.21}\)+......+\(\frac{5}{61.66}\) )
D = \(\frac{3}{5}\). ( \(\frac{1}{11}\)- \(\frac{1}{16}\) + \(\frac{1}{16}\)- \(\frac{1}{21}\)+ ......... + \(\frac{1}{61}\)- \(\frac{1}{66}\))
D =\(\frac{3}{5}\). ( \(\frac{1}{11}\)- \(\frac{1}{66}\))
D = \(\frac{3}{5}\). \(\frac{5}{66}\)
D = \(\frac{1}{22}\)
# HOK TỐT #
11 tháng 4 2019
\(S1=2+4+6+...+150=\frac{2+150}{2}\cdot\left(\frac{150-2}{2}+1\right)\)
\(S1=\frac{152}{2}\cdot\left(\frac{148}{2}+1\right)=76\cdot\frac{150}{2}=76\cdot75=5700\)
- S3 và S5 tương tự nha bạn :vv
\(S2=5^2+5^3+5^4+...+5^{100}\)
\(\Rightarrow5S2=5^3+5^4+5^5+...+5^{100}+5^{101}\)
\(S2=5^2+5^3+5^4+5^5+...+5^{100}\)
\(\Rightarrow5S2-S2=4S2=5^{101}-5^2\Rightarrow S2=\frac{5^{101}-5^2}{4}\)
\(S4=\frac{5}{11\cdot16}+\frac{5}{16\cdot21}+...+\frac{5}{61\cdot66}\)
\(\Rightarrow S4=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)
\(\Rightarrow S4=\frac{1}{11}-\frac{1}{16}=\frac{16}{176}-\frac{11}{176}=\frac{5}{176}\)
DT
14 tháng 7 2016
\(A=\)\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{51.56}\)
\(5A=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{51.56}\)
\(5A=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{51}-\frac{1}{56}\)
\(5A=1-\frac{1}{56}=\frac{55}{56}\)
\(A=\frac{55}{56}\div5=\frac{55}{56}.\frac{1}{5}=\frac{11}{56}\)
LT
1
22 tháng 7 2019
a) Đề phải là thế này chứ \(\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{101.106}\)
Giai
\(=\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{101.106}\)
\(=\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{101}-\frac{1}{106}\)
\(=\frac{1}{6}-\frac{1}{106}\)
\(=\frac{25}{159}\)
b) Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{100}}\)
\(\Rightarrow5A=1+\frac{1}{5}+...+\frac{1}{5^{99}}\)
\(\Rightarrow5A-A=\left(1+\frac{1}{5}+...+\frac{1}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{100}}\right)\)
\(\Rightarrow4A=1-\frac{1}{5^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{5^{100}}}{4}\)
TN
16 tháng 4 2016
đặt A=100^10+1/100^10-1
B=10^100+1/10^100-3
ta có:\(A=\frac{100^{10}+1}{100^{10}-1}=\frac{100^{10}-1+2}{100^{10}-1}=\frac{100^{10}-1}{100^{10}-1}+\frac{2}{100^{10}-1}=1+\frac{2}{100^{10}-1}\)
\(B=\frac{10^{100}+1}{10^{100}-3}=\frac{10^{100}-3+4}{10^{100}-3}=\frac{10^{100}-3}{10^{100}-3}+\frac{4}{10^{100}-3}=1+\frac{4}{10^{100}-3}=1+\frac{4}{100^{10}-3}\)
vì 10010-1>10010-3
=>\(\frac{4}{100^{10}-1}<\frac{4}{100^{10}-3}\)
=>A<B
D
6 tháng 3 2016
\(\frac{5x}{1.6}+\frac{5x}{6.11}+\frac{5x}{11.16}+\frac{5x}{16.21}=\frac{1}{25}\)
\(x\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}\right)=\frac{1}{25}\)
\(x\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}\right)=\frac{1}{25}\)
\(x\left(1-\frac{1}{21}\right)=\frac{1}{25}\)
\(\frac{20}{21}x=\frac{1}{25}\)
\(x=\frac{1}{25}:\frac{20}{21}=.....\)
LH
4 tháng 10 2018
\(A=1+5+5^2+5^3+..+5^{100}\)
\(5A=5+5^2+5^3+..+5^{101}\)
\(A=\frac{5^{101}-1}{4}\)\(SUYRA\) \(A< B\)
4 tháng 10 2018
\(A=5^0+5+5^2+...+5^{100}.\)
\(\Rightarrow5A=5+5^2+5^3+...+5^{101}\)
\(\Rightarrow5A-A=4A=\left(5+5^2+5^3+...+5^{101}\right)-\left(5^0+5+5^2+...+5^{100}\right)\)
\(=5^{101}-1\)
\(\Rightarrow A=\frac{5^{101}-1}{4}\)
Còn lại tự lm nha bn
giúp mk thành bài giải nha
\(=21\left(16+84\right)-100=21\cdot100-100=100\left(21-1\right)=100\cdot20=2000\)