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a, Ta có : 22013 = (23)671 = 8671
31344 = (32)672 = 9672
Mà 8671 < 9672
Vậy 22013 < 31344
b, \(A=\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{64\cdot69}\)
\(A\cdot5=\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+...+\frac{5}{64\cdot69}\)
\(A\cdot5=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{64}-\frac{1}{69}\)
\(A\cdot5=\frac{1}{4}-\frac{1}{69}=\frac{65}{276}\)
\(A=\frac{65}{276}\div5=\frac{13}{276}\)
a, Ta có: 22013 = (23)671 = 8671
31344 = (32)672 = 9672
Vì 8671 < 9672 => 22013 < 31344
b, A = \(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{64.69}\)
5A = \(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{64.69}\)
5A = \(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{64}-\frac{1}{69}\)
5A = \(\frac{1}{4}-\frac{1}{69}=\frac{65}{276}\)
A = \(\frac{65}{276}:5=\frac{13}{276}\)
5B=\(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+...+\frac{5}{64\cdot69}\)
5B=\(\frac{9-4}{4\cdot9}+\frac{14-9}{9\cdot14}+...+\frac{69-64}{64.69}\)
5B=\(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{64}-\frac{1}{69}\)
5B=\(\frac{65}{276}\)
B=\(\frac{13}{276}\)
\(B=\frac{1}{4.9}+\frac{1}{9.14}+....+\frac{1}{64.69}\)
\(\Rightarrow5B=\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{64.69}\)
\(5B=\frac{9-4}{4.9}+\frac{14-9}{9.14}+....+\frac{69-64}{64.69}\)
\(5B=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{64}-\frac{1}{69}\)
\(5B=\frac{1}{4}-\frac{1}{69}\)
\(5B=\frac{65}{276}\)
\(B=\frac{65}{276}:5\)
\(B=\frac{13}{276}\)
a 2^2015>3^1029
b 5A=\(\frac{5}{4.9}\)+\(\frac{5}{9.14}\)+\(\frac{5}{14.19}\)+.....+\(\frac{5}{64.69}\)
5A=1/4-1/9+1/9-1/14+1/14-1/19+1/19+....+1/64-1/69
5A=1/4-1/9
A=(1/4-1/9)/5
A=1/36
\(D=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2017}{2018}\)
\(D=\frac{1}{2018}\)
Vậy \(D=\frac{1}{2018}\)
\(E=\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{64.69}+\frac{1}{69.74}\)
\(E=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{69}-\frac{1}{74}\right)\)
\(E=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{74}\right)\)
\(E=\frac{1}{5}\cdot\frac{35}{148}=\frac{7}{148}\)
Vậy E = ...
= \(\frac{1}{4}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{14}\)+\(\frac{1}{14}\)-\(\frac{1}{19}\)+... + \(\frac{1}{44}\)-\(\frac{1}{49}\)
= \(\frac{1}{4}\)-\(\frac{1}{49}\)
= \(\frac{45}{196}\)
ai tốt bụng thì tk cho mk nha, mk đg âm điểm nè huhu
b: \(C=75\left(2-128+128\right)=75\cdot2=150\)
e: \(E=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{69\cdot74}\right)\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{69}-\dfrac{1}{74}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{70}{74}=\dfrac{14}{74}=\dfrac{7}{37}\)
\(A=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{65}{276}=\dfrac{13}{276}\)