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Ta có: 527=(53)9=1259<1289=(27)9=263
=>527<263(1)
Lại có: 263<264=(216)4=655364<781254=(57)4=528
=>263<264<528
=>263<528(2)
Từ (1) và (2) ta thấy:
527<263<528
=>ĐPCM
Ta có :
527 = ( 53 )9 = 1259 < 1289 = ( 27 )9 = 263
=> 527 < 263
Mà 263 < 264 = ( 216 )4 = 655364 < 528 = ( 57 )4 = 781254
=> 527 < 263 < 528
a)A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) > 1 / (1*2) + 1 / (3*4) = 1 / 2 + 1 / 12 = 7 / 12 ♦
A = 1 / (1*2) + 1 / (3*4) + ... + 1 / (99*100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100) =
(1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 <
1 - 1 / 2 + 1 / 3 = 5 / 6 ♥
♦, ♥ => 7 / 12 < A < 5 / 6
b)ta có:
1/1.2+1/3.4+1/5.6+...+1/49.50
=>1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50
=>(1+1/3+1/5+1/7+...+1/49)-(1/2+1/4+1/6+...+1/50)
=>(1+1/2+1/3+...+1/49+1/50)-(1/2+1/4+1/6+...+1/50).2
=>(1+1/2+1/3+...+1/49+1/50) -( 1+1/2+1/3+...+1/25)
=>1/26+1/27+1/28+...+1/50=1/26+1/27+1/28+...+1/50
hay 1/1.2+1/3.4+1/5.6+...+1/49.50=1/26+1/27+1/28+...+1/50
\(\dfrac{4}{15}-\dfrac{23}{28}-\left(-\dfrac{23}{28}+-\dfrac{11}{5}-\dfrac{29}{27}\right)-\dfrac{2}{27}\)
\(=\dfrac{4}{15}-\dfrac{23}{28}+\dfrac{23}{28}+\dfrac{11}{5}+\dfrac{29}{27}-\dfrac{2}{27}\)
\(=\left(\dfrac{4}{15}+\dfrac{11}{5}\right)+\left(-\dfrac{23}{28}+\dfrac{23}{28}\right)+\left(\dfrac{29}{27}-\dfrac{2}{27}\right)\)
\(=\left(\dfrac{4}{15}+\dfrac{33}{15}\right)+0+1\)
\(=\dfrac{37}{15}+1\)
\(=\dfrac{52}{15}\)
Ta có :
\(5^{27}=\left(5^3\right)^9=125^9