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A = 22017 - 22016 - 22015 - … - 25
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H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
a) \(A=1+2+2^2+...+2^{80}\)
\(2A=2+2^2+2^3+...+2^{81}\)
\(2A-A=2+2^2+2^3+...+2^{81}-1-2-2^2-...-2^{80}\)
\(A=2^{81}-1\)
Nên A + 1 là:
\(A+1=2^{81}-1+1=2^{81}\)
b) \(B=1+3+3^2+...+3^{99}\)
\(3B=3+3^2+3^3+...+3^{100}\)
\(3B-B=3+3^2+3^3+...+3^{100}-1-3-3^2-...-3^{99}\)
\(2B=3^{100}-1\)
Nên 2B + 1 là:
\(2B+1=3^{100}-1+1=3^{100}\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
2)
a) \(2^x\cdot\left(1+2+2^2+...+2^{2015}\right)+1=2^{2016}\)
Gọi:
\(A=1+2+2^2+...+2^{2015}\)
\(2A=2+2^2+2^3+...+2^{2016}\)
\(A=2^{2016}-1\)
Ta có:
\(2^x\cdot\left(2^{2016}-1\right)+1=2^{2016}\)
\(\Rightarrow2^x\cdot\left(2^{2016}-1\right)=2^{2016}-1\)
\(\Rightarrow2^x=\dfrac{2^{2016}-1}{2^{2016}-1}=1\)
\(\Rightarrow2^x=2^0\)
\(\Rightarrow x=0\)
b) \(8^x-1=1+2+2^2+...+2^{2015}\)
Gọi: \(B=1+2+2^2+...+2^{2015}\)
\(2B=2+2^2+2^3+...+2^{2016}\)
\(B=2^{2016}-1\)
Ta có:
\(8^x-1=2^{2016}-1\)
\(\Rightarrow\left(2^3\right)^x-1=2^{2016}-1\)
\(\Rightarrow2^{3x}-1=2^{2016}-1\)
\(\Rightarrow2^{3x}=2^{2016}\)
\(\Rightarrow3x=2016\)
\(\Rightarrow x=\dfrac{2016}{3}\)
\(\Rightarrow x=672\)
NX
15 tháng 8 2023
Ta có:
A = 2 + 22 + 23 + … + 22017
2A = 2.( 2 + 22 + 23 + … + 22017)
2A = 22 + 23 + 24 + … + 22018
2A – A = (22 + 23 + 24 + … + 22018) – (2 + 22 + 23 + … + 22017)
Vậy A = 22018 – 2
NX
16 tháng 8 2023
Ta có: A = 2 + 22 + 23 + … + 22017
2A = 2.( 2 + 22 + 23 + … + 22017)
2A = 22 + 23 + 24 + … + 22018
2A – A = (22 + 23 + 24 + … + 22018) – (2 + 22 + 23 + … + 22017)
A = 22018 – 2
Vậy A = 22018 – 2
DX
1
AH
Akai Haruma
Giáo viên
13 tháng 3 2021
Lời giải:
Xét tử số:
$\text{TS}=1+25^4+25^8+...+25^{28}$
$25^4.\text{TS}=25^4+25^8+...+25^{32}$
$\Rightarrow \text{TS}(25^4-1)=25^{32}-1$
$\text{TS}=\frac{25^{32}-1}{25^4-1}$
Xét mẫu số:
$\text{MS}=1+25^2+..+25^{30}$
$25^2.\text{MS}=25^2+25^4+...+25^{32}$
$\Rightarrow \text{MS}(25^2-1)=25^{32}-1$
$\Rightarrow \text{MS}=\frac{25^{32}-1}{25^2-1}$
Do đó:
$A=\frac{25^{32}-1}{25^4-1}:\frac{25^{32}-1}{25^2-1}=\frac{25^2-1}{25^4-1}$
$=\frac{25^2-1}{(25^2-1)(25^2+1)}=\frac{1}{25^2+1}$
TD
2
S
20 tháng 2 2023
a)\(\dfrac{22}{25}+\dfrac{25}{125}+\dfrac{36}{96}\)
\(=\dfrac{22}{25}+\dfrac{1}{5}+\dfrac{3}{8}\)
\(=\dfrac{176}{200}+\dfrac{40}{200}+\dfrac{75}{200}\)
\(=\dfrac{291}{200}\)
b)\(\dfrac{22}{77}+\dfrac{56}{98}+\dfrac{25}{105}\)
\(=\dfrac{2}{7}+\dfrac{4}{7}+\dfrac{5}{21}\)
\(=\dfrac{6}{7}+\dfrac{5}{21}\)
\(=\dfrac{18}{21}+\dfrac{5}{21}=\dfrac{23}{21}\)
CM
2 tháng 10 2019
Đáp án C
A = 2 5 + 2 7 + 2 17 - 2 395 3 5 + 3 7 + 3 17 - 3 395 = 2 . 1 5 + 2 . 1 7 + 2 . 17 - 2 1 395 3 . 1 5 + 3 . 1 7 + 3 . 1 17 - 3 . 1 195 = 2 1 5 + 1 7 + 1 17 - 1 395 3 1 5 + 1 7 + 1 17 - 1 395 = 2 3
\(A=2^{2017}-2^{2016}-2^{2015}-..........-2^5\)
\(\Leftrightarrow A=2^{2017}-\left(2^{2016}+2^{2015}+..........+2^5\right)\)
Đặt :
\(B=2^{2016}+2^{2017}+...........+2^5\)
\(\Leftrightarrow2B=2^{2017}+2^{2016}+..........+2^6\)
\(\Leftrightarrow2B-B=\left(2^{2017}+2^{2016}+.......+2^6\right)-\left(2^{2016}+2^{2015}+......+2^5\right)\)
\(\Leftrightarrow B=2^{2017}-2^5\)
\(\Leftrightarrow A=2^{2017}-\left(2^{2017}-2^5\right)\)
\(\Leftrightarrow A=2^{2017}-2^{2017}-2^5\)
\(\Leftrightarrow A=0+2^5\)
\(\Leftrightarrow A=32\)
A = 22017 - 22016 - 22015 - … - 25
= 22017 - (22016 + 22015 + … + 25)
Đặt E = 22016 + 22015 + … + 25
2E = 22017 + 22016 + … + 26
2E - E =(22017 - 22016 - … - 26) - (22016 - 22015 - … - 25)
E = 22017 - 25
=> A = 22017 - (22017 - 25)
= 22017 - 22017 + 25
= 32