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H24
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NT
11 tháng 4 2017 lúc 21:58

\(\left(\dfrac{1}{64}-\dfrac{1}{3^2}\right)\left(\dfrac{1}{64}-\dfrac{1}{4^2}\right)...\left(\dfrac{1}{64}-\dfrac{1}{64^2}\right)\)

\(=\left(\dfrac{1}{64}-\dfrac{1}{3^2}\right)\left(\dfrac{1}{64}-\dfrac{1}{4^2}\right)...\left(\dfrac{1}{64}-\dfrac{1}{8^2}\right)...\left(\dfrac{1}{64}-\dfrac{1}{64^2}\right)\)

\(=\left(\dfrac{1}{64}-\dfrac{1}{3^2}\right)\left(\dfrac{1}{64}-\dfrac{1}{4^2}\right)...0...\left(\dfrac{1}{64}-\dfrac{1}{64^2}\right)\)

\(=0\)

Vậy...

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TN
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HH
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NV
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NT
7 tháng 6 2022 lúc 22:59

\(=\left(\dfrac{1}{3}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}\right)+\left(\dfrac{3}{5}+\dfrac{1}{5}\right)+\dfrac{1}{64}\)

\(=\dfrac{12-9-8-1}{36}+\dfrac{4}{5}+\dfrac{1}{64}\)

\(=\dfrac{-1}{6}+\dfrac{1}{64}+\dfrac{4}{5}=\dfrac{623}{960}\)

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H24
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DH
22 tháng 6 2021 lúc 23:47

\(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)

\(=\frac{1}{2}\left(5^{32}-1\right)\left(5^{32}+1\right)\)

\(=\frac{1}{2}\left(5^{64}-1\right)\)

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LC
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VP
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TL
22 tháng 5 2021 lúc 21:45

a) `1/2 xx 5/4 + 1/2 xx 3/4 + 1/2`

`=1/2 xx (5/4 + 3/4 + 1)`

`=1/2 xx 3`

`=3/2`

b) `72/95 xx 64/72 xx 95/72 xx 14/86 xx 72/64 xx 86/14`

`=(72xx64xx95xx14xx72xx86)/(95xx72xx72xx86xx64xx14)`

`= 1`.

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a)1/2.(5/4+3/4+1)

=1/2.3

=3/2

 

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b)=(72/95.95/72).(64/72.72/64).(14/86.86/14)

=1.1.1=1

 

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NH
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MB
20 tháng 4 2019 lúc 20:12

Ta có: \(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{64^2}< \frac{1}{4^2}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{63.64}\)

\(\frac{1}{4^2}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{63.64}=\frac{1}{4^2}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{63}-\frac{1}{64}\)

\(=\frac{1}{4^2}+\frac{1}{4}-\frac{1}{64}\)

VÌ: \(\frac{1}{4^2}+\frac{1}{4}-\frac{1}{64}< \frac{1}{4^2}+\frac{1}{4}=\frac{5}{16}\)

Nên: \(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{64^2}< \frac{5}{16}\left(dpcm\right)\)

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NN
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KL
14 tháng 7 2016 lúc 22:05

a) (2+1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2+1)(2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2^2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2^4-1)(2^4+1)....(2^32+1)-2^64

=......

=(2^32-1)(2^32+1)-2^64

=2^64-1-2^64=-1

b)Đặt A=(5+3)(5^2+3^2)(5^4+3^4)...(5^64+3^64)+(5^128-3^128)/2

đặt B=(5+3)(5^2+3^2)(5^4+3^4)...(5^64+3^64)

\(2B=\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=\left(5^4-3^4\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=.......\)

2B=(5^64-3^64)(5^64+3^64)

2B=5^128-3^128

B=(5^128-3^128)/2 (thế vào đề bài)

=> A=B+(5^128-3^128)/2=(5^128-3^128)/2+(5^128-3^128)/2=\(\frac{2\left(5^{128}-3^{128}\right)}{2}=\left(5^{128}-3^{128}\right)\)

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DH
14 tháng 7 2016 lúc 21:29

a) A = ( 2-1)(2+1)(22+1)...(232+1)-264

         =(22-1)(22+1)(24+1)... -264

       =....

       =264-1-264=1

câu b tương tự nhá

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