Bây giờ cậu cần không thế;D

h mik ko gấp nữa, nhưng nếu cậu biết cách giải thì chỉ mik nha ạ, làm tư liệu sau này mik học ý ạ :>

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 = 2^x

⇒ 2²⁰²³ - 2 + 2 = 2^x

⇒ 2²⁰²³ = 2^x

Vậy x = 2023

Lại có : 4B + 5 = 5^x

⇒ 4 . \(\dfrac{5^{2023}-5}{4}\) + 5 = 5^x

⇒ 5²⁰²³ - 5 + 5 = 5^x

⇒ 5²⁰²³ = 5^x

Vậy x = 2023

a)=2012

b)=12134

c)= tự suy nghĩ

blêu blêu

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) \(A=1+2+2^2+...+2^{80}\)

\(2A=2+2^2+2^3+...+2^{81}\)

\(2A-A=2+2^2+2^3+...+2^{81}-1-2-2^2-...-2^{80}\)

\(A=2^{81}-1\)

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:

\(8^x-1=2^{2016}-1\)

\(\Rightarrow\left(2^3\right)^x-1=2^{2016}-1\)

\(\Rightarrow2^{3x}-1=2^{2016}-1\)

\(\Rightarrow2^{3x}=2^{2016}\)

\(\Rightarrow3x=2016\)

\(\Rightarrow x=\dfrac{2016}{3}\)

\(\Rightarrow x=672\)

1) \(B\left(24\right)=\left\{24;48;72;96\right\}\)

\(B\left(39\right)=\left\{39;78\right\}\)

2) a) \(x+20⋮x+2\)

\(\Rightarrow x+20-\left(x+2\right)⋮x+2\)

\(\Rightarrow x+20-x-2⋮x+2\)

\(\Rightarrow18⋮x+2\)

\(\Rightarrow x+2\in\left\{1;2;3;6;9;18\right\}\)

\(\Rightarrow x\in\left\{-1;0;1;4;7;16\right\}\)

\(\Rightarrow x\in\left\{0;1;4;7;16\right\}\left(x\in N\right)\)

b) \(x+5⋮4x+69\)

\(\Rightarrow4\left(x+5\right)-\left(4x+69\right)⋮4x+69\)

\(\Rightarrow4x+20-4x-69⋮4x+69\)

\(\Rightarrow-49⋮4x+69\)

\(\Rightarrow4x+69\in\left\{1;7;49\right\}\)

\(\Rightarrow x\in\left\{-17;-\dfrac{31}{2};-20\right\}\)

\(\Rightarrow x\in\varnothing\left(x\in N\right)\)

c) \(10x+23⋮2x+1\)

\(\Rightarrow10x+23-5\left(2x+1\right)⋮2x+1\)

\(\Rightarrow10x+23-10x-5⋮2x+1\)

\(\Rightarrow18⋮2x+1\)

\(\Rightarrow2x+1\in\left\{1;2;3;6;9;18\right\}\)

\(\Rightarrow x\in\left\{0;\dfrac{1}{2};1;\dfrac{5}{2};4;\dfrac{17}{2}\right\}\)

\(\Rightarrow x\in\left\{0;1;4\right\}\left(x\in N\right)\)

Đính chính câu 1

Không có số có 2 chữ số thỏa đề bài