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3 tháng 1 2018
B=2^10*(2*3)^15+3^14*3*5*(2^2)^13/2^19*(2*3^2)^7*3-3^15*2^25
B=2^10*2^15*3^15+3^15+3^15*5*2^26/2^19*2^7*3^14**3-3^15*2^25
B=2^25*3^15+5*2*3^15*2^25/2*2^25*3^15-3^15*2^25
B=1*2^25*3^15+10*2^25*3^15/2*2^25*3^15-2^25*3^15
B=2^25*3^15*(1+10)/2^25*3^15*(2-1)=11/1=11
- B=2^10*6^15+3^14*15*4^13/2^19*18^7*3-3^15*2^25
DH
1
13 tháng 5 2015
1:
I2x+3I = 5
=> 2x+3 = 5 hoặc 2x+3 = -5
=> 2x = 5 - 3 hoặc 2x = -5 - 3
=> 2x = 2 hoặc 2x = -8
=> x = 2 hoặc x = -4
2:
B = 1/2.2/3.3/4.4/5.....27/28
= 1.2.3.4.5.6...27/2.3.4.5.6...28
= 1/28
3:
2A = 2(1+1/2+1/2^2+1/2^3+1/2^4+...+1/2^2015) = 2+1+1/2+1/2^2+1/2^3+...+1/2^2014
=> 2A-A = ( 2+1+1/2+1/2^2+1/2^3+...+1/2^2014)-(1+1/2+1/2^2+1/2^3+...+1/2^2015)
=> A = 2-1/2^2015
18 tháng 9 2018
a=\(1+2+2^2+..+2^{25}\)(1)
2a=\(2+2^2+2^3+...+2^{26}\)(2)
trừ vế với vế của 2 cho 1
2a-a =\(\left(2+2^2+..+2^{26}\right)-\left(1+2+..+2^{25}\right)\)
a=\(2^{26}-1\)
b a=\(1+2+...+2^{25}\)
a=\(\left(1+2\right)+\left(2^2+2^3\right)...+\left(2^{24}+2^{25}\right)\)
a=3+\(2^2.\left(1+2\right)\).......+\(2^{24}.\left(1+2\right)\)
a=3+\(2^2.3\)+....+\(2^{24}.3\)
a=3.(\(1+2^2+...+2^{24}\))\(⋮\)3
=>đpcm
LH
18 tháng 9 2018
1, A = 1 + 2 + 22 + ... + 225
2A = 2 + 22 + 23 + ... + 226
2A - A = ( 2 + 22 + 23 + ... + 226 ) - ( 1 + 2 + 22 + ... + 225 )
A = 226 - 1
Vậy A = 226 - 1
2, A = 1 + 2 + 22 + ... + 225
A = ( 1 + 2 ) + ( 22 + 23 ) + ... + ( 224 + 225 )
A = 3 + 22 ( 1 + 2 ) + ... + 224 ( 1 + 2 )
A = 3 ( 1 + 22 + ... + 224 ) \(⋮\)3
Vậy A \(⋮\)3
Hok tốt
NV
1
13 tháng 8 2017
\(\dfrac{-2}{1.3}-\dfrac{2}{3.5}-\dfrac{2}{5.7}-...-\dfrac{2}{19.21}-\dfrac{2}{21.23}-\dfrac{2}{23.25}-\dfrac{2}{25.27}-\dfrac{1}{27}\)\(=\dfrac{-2}{1.3}+\dfrac{-2}{3.5}+\dfrac{-2}{5.7}+...+\dfrac{-2}{19.21}+\dfrac{-2}{21.23}+\dfrac{-2}{23.25}+-\dfrac{2}{25.27}-\dfrac{1}{27}\)Đặt:
\(A=\dfrac{-2}{1.3}+\dfrac{-2}{3.5}+\dfrac{-2}{5.7}+...+\dfrac{-2}{19.21}+\dfrac{-2}{21.23}+\dfrac{-2}{23.25}+\dfrac{-2}{25.27}\)\(A=-1\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{25}-\dfrac{1}{27}\right)\)
\(A=-1\left(1-\dfrac{1}{27}\right)\)
\(A=-1.\dfrac{26}{27}=-\dfrac{26}{27}\)
Thay vào biểu thức ban đầu ta có:
Giá trị là:
\(-\dfrac{26}{27}-\dfrac{1}{27}=-\dfrac{27}{27}=-1\)
DD
Đoàn Đức Hà
Giáo viên
25 tháng 10 2021
a) \(A=1+3+3^2+...+3^{100}\)
\(3A=3+3^2+3^3+...+3^{101}\)
\(3A-A=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)
\(2A=3^{101}-1\)
\(A=\frac{3^{101}-1}{2}\)
b) \(B=2^{100}-2^{99}+2^{98}-2^{97}+...-2^3+2^2-2+1\)
\(2B=2^{101}-2^{100}+2^{99}-2^{98}+...-2^4+2^3-2^2+2\)
\(B+2B=\left(2^{100}-2^{99}+...-2+1\right)+\left(2^{101}-2^{100}+...-2^2+2\right)\)
\(3B=2^{101}+1\)
\(B=\frac{2^{101}+1}{3}\)
3 tháng 2 2019
a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012
2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013
3M=2^0+2^2013
M=(2^0+2^2013)÷3
Vậy.......
b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012
3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013
4N=3-3^2013
N=(3-3^2013)÷4
Vậy........
K tao nhé ko lên lớp tao đánh m😈😈😈
2 tháng 2 2021
a) Ta có: \(\dfrac{25^{28}+25^{24}+25^{20}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}\)
\(=\dfrac{25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+...+\left(25^4+1\right)}{25^{28}\left(25^2+1\right)+25^{24}\left(25^2+1\right)+...+\left(25^2+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\cdot\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}{\left(25^2+1\right)\left[25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+25^8\left(25^4+1\right)+\left(25^4+1\right)\right]}\)
\(=\dfrac{\left(25^4+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}\)
\(=\dfrac{1}{25^2+1}=\dfrac{1}{626}\)
\(A=1+2+2^2+...+2^{25}\)
\(2A=2+2^2+2^3+...+2^{26}\)
\(2A-A=A=2^{26}-1\)
\(B=1+3+3^2+...+3^{19}\)
\(3B=3+3^2+3^3+...+3^{20}\)
\(3B-B=3^{20}-1\)
\(B=\dfrac{3^{20}-1}{2}\)
a: A=1+2+2^2+...+2^25
=>2A=2+2^2+...+2^26
=>A=2^26-1
b: B=1+3+3^2+...+3^19
=>3B=3+3^2+3^3+...+3^20
=>2B=3^20-1
=>B=(3^20-1)/2