1, cho A = 2 + 22 + 23 + ....+ 22017
a) Tính A
b) Tìm x, biết : A + 2 = 2x
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a) Ta có A = 21 + 22 + 23 + ... + 22022
2A = 22 + 23 + 24 + ... + 22023
2A - A = ( 22 + 23 + 24 + ... + 22023 ) - ( 21 + 22 + 23 + ... + 22022 )
A = 22023 - 2
Lại có B = 5 + 52 + 53 + ... + 52022
5B = 52 + 53 + 54 + ... + 52023
5B - B = ( 52 + 53 + 54 + ... + 52023 ) - ( 5 + 52 + 53 + ... + 52022 )
4B = 52023 - 5
B = \(\dfrac{5^{2023}-5}{4}\)
b) Ta có : A + 2 = 2x
⇒ 22023 - 2 + 2 = 2x
⇒ 22023 = 2x
Vậy x = 2023
Lại có : 4B + 5 = 5x
⇒ 4 . \(\dfrac{5^{2023}-5}{4}\) + 5 = 5x
⇒ 52023 - 5 + 5 = 5x
⇒ 52023 = 5x
Vậy x = 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:
\(2B+1=3^{100}-1+1=3^{100}\)
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:
\(8^x-1=2^{2016}-1\)
\(\Rightarrow\left(2^3\right)^x-1=2^{2016}-1\)
\(\Rightarrow2^{3x}-1=2^{2016}-1\)
\(\Rightarrow2^{3x}=2^{2016}\)
\(\Rightarrow3x=2016\)
\(\Rightarrow x=\dfrac{2016}{3}\)
\(\Rightarrow x=672\)
a) \(A=2+2^2+2^3+...+2^{2022}\)
\(2A=2.\left(2+2^2+2^3+...+2^{2022}\right)\)
\(2.A=2^2+2^3+2^4+...+2^{2023}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2023}\right)-\left(2+2^2+2^3+...+2^{2022}\right)\)
\(A=2^{2023}-2\)
b) A + 2 = 2x
Hay \(\left(2^{2023}-2\right)+2=2^x\)
\(2^{2023}-2+2=2^x\)
\(2^{2023}=2^x\)
\(\Rightarrow x=2023\)
a, A = 21 + 22 + 23 + ...+ 22022
2A = 22 + 23 +...+ 22022 + 22023
2A - A = 22023 - 21
A = 22023 - 2
b, A + 2 = 2\(^x\) ⇒ 22023 - 2 + 2 = 2\(x\)
22023 = 2\(^x\)
2023 = \(x\)
a, 2.(x – 5)+7 = 77
<=> 2.(x – 5) = 70 <=> x – 5 = 35 <=> x = 40
b, x - 1 3 - 3 5 : 3 4 + 2 . 2 3 = 14
<=> x - 1 3 - 3 + 2 4 = 14
<=> x - 1 3 = 14 + 3 - 16 = 1
<=> x – 1 = 1 <=> x = 2
c, 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 = 2 x - 1 - 1
Đặt: A = 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 => 2A = 2 + 2 2 + 2 3 + . . . + 2 2017
=> 2A – A = ( 2 + 2 2 + 2 3 + . . . + 2 2017 ) – ( 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 )
=> A = 2 2017 - 1
Ta có: 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 = 2 x - 1 - 1 => 2 2017 - 1 = 2 x - 1 - 1 => x = 2018
d, 5 2 x - 3 - 2 . 5 2 = 5 2 . 3
<=> 5 2 x - 3 = 5 2 . 3 + 5 2 . 2
<=> 5 2 x - 3 = 5 2 . ( 3 + 2 )
<=> 5 2 x - 3 = 5 3
<=> 2x – 3 = 3 => x = 3
a) 22 + (2x -13) = 83 => 2x -13 = 61 => x = 37.
b) 51 - (-12 + 3x) = 27 => 63 - 3x = 27 => x = 12.
c) - (2x + 2) + 21 = - 23 => 2x + 2 = 44 => x = 21.
d) 25 - (25 - x) = 0 => 25 - 25 + x = 0 => x = 0.
a: Tổng các số hạng là:
\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)
Ta có: A+1=2x
\(\Leftrightarrow2x=24311\)
hay \(x=\dfrac{24311}{2}\)
Bài 2 :
a, \(x=\dfrac{3}{5}-\dfrac{7}{8}=\dfrac{24-30}{40}=-\dfrac{6}{40}=-\dfrac{3}{20}\)
b, \(2x-1=-2\Leftrightarrow x=-\dfrac{1}{2}\)
a)A= 2+22+23+.....+22017
A*2=22+23+24+......+22017+22018
a*2-a=(22+23+24+......+22017+22018)-(2+22+23+......+22017)
a*1=22018-2
b)A+2=2X
22018-2+2=2X
=>2X=22018
X=2018
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