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Ta có : x = 1 + 2 + 2² + ... + 2¹⁰⁰

=> 2x = 2 + 2² + 2³ + ... + 2¹⁰¹

Lấy 2x trừ x theo vế ta có :

2x - x = (2 + 2² + 2³ + ... + 2¹⁰¹) - (1 + 2 + 2² + ... + 2¹⁰⁰)

       x = 2¹⁰¹ - 1

mà y - x = 2¹⁰¹ - (2¹⁰¹ - 1)

              = 2¹⁰¹ - 2¹⁰¹ + 1

              = 1

=> x;y là 2 số tự nhiên liên tiếp (đpcm)

\(A=3+3^2+....+3^{99}\)

\(3A=3^2+3^3+...+3^{100}\)

\(3A-A=3^2+3^3+...+3^{100}-3-3^2-...-3^{99}\)

\(2A=3^{100}-3\)

\(A=\dfrac{3^{100}-3}{2}\)

\(\Rightarrow2A+3=9^{2x+6}\)

\(\Rightarrow2\cdot\dfrac{3^{100}-3}{2}+3=\left(3^2\right)^{2x+6}\)

\(\Rightarrow3^{100}-3+3=3^{2\left(2x+6\right)}\)

\(\Rightarrow3^{100}=3^{4x+12}\)

\(\Rightarrow4x+12=100\)

\(\Rightarrow4x=88\)

\(\Rightarrow x=22\)

\(A=3+3^2+3^3+\cdots+3^{99}\)

=>\(3A=3^2+3^3+3^4+\cdots+3^{100}\)

=>\(3A-A=3^2+3^3+3^4+\cdots+3^{100}-3-3^2-\cdots-3^{99}\)

=>\(2A=3^{100}-3\)

=>\(2A+3=3^{100}\)

=>\(9^{2x+6}=3^{100}=\left(3^2\right)^{50}=9^{50}\)

=>2x+6=50

=>2x=44

=>\(x=\frac{44}{2}=22\)

a)   \(3^{x.3}=3^5\)

=> x.3=5

<=>x=5/3

b) x^3=8^2

<=> x^3=(2^3)2

<=>x=2^2

<=>x=4

c) x^20=x

x=1 hoac =0  

chúc bn học tốt 

kkkkk

\(x^3=343\)

hay x=7

Lời giải:

$1+2+2^3+2^4+2^5+...+2^{x+1}=1023$

$2^3+2^4+2^5+...+2^{x+1}=1020(1)$

$2^4+2^5+2^6+...+2^{x+2}=2040(2)$

Lấy (2) trừ (1) theo vế suy ra:

$2^{x+2}-2^3=2040-1020=1020$

$2^{x+2}=1028$

Với giá trị này sẽ không tồn tại số tự nhiên x. Bạn xem lại đề.

1. 15^x=225

   <=>15^x=15^2

    =>x=2

    vậy....

2. x^3=27

   <=>x^3=3^3

   =>x=3

vậy......

3. 4.2^x-3=125

   4.2^x=128

   2^x=32

   2^x=2^5

   =>x=5

vậy.....

\(15^x=225\)

\(\Leftrightarrow15^x=15^2\)

\(\Leftrightarrow x=2\)

\(x^3=27\)

\(\Leftrightarrow x^3=3^3\)

\(\Leftrightarrow x=3\)

\(4.2^x-3=125\)

\(\Leftrightarrow4.2^x=125+3\)

\(\Leftrightarrow4.2^x=128\)

\(\Leftrightarrow2^x=128:4\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Leftrightarrow x=5\)

~ Hok tốt ~

Câu a mình ko bt trình bày thông cảm

b) \(^{2^x.\left(2^2\right)^2=\left(2^3\right)^2}\)

\(2^x.2^4=2^6\)

\(2^x=2^6:2^4\)

\(2^x=2^2\)

\(x=2\)

a)(x^5)^10=x<=>x=0 hoặc 1

a) \(5\left(x+7\right)-10=2^3\cdot5\)

\(\Rightarrow5\left(x+7\right)-10=40\)

\(\Rightarrow5\left(x+7\right)=40+10\)

\(\Rightarrow x+7=\dfrac{50}{5}\)

\(\Rightarrow x+7=10\)

\(\Rightarrow x=10-7\)

\(\Rightarrow x=3\)

b) \(9x-2\cdot3^2=3^4\)

\(\Rightarrow9x-18=81\)

\(\Rightarrow9x=81+18\)

\(\Rightarrow9x=99\)

\(\Rightarrow x=\dfrac{99}{9}\)

\(\Rightarrow x=11\)

c) \(5^{25}\cdot5^{x-1}=5^{25}\)

\(\Rightarrow5^{x-1}=5^{25}:5^{25}\)

\(\Rightarrow5^{x-1}=1\)

\(\Rightarrow5^{x-1}=5^0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

a) 5(�+7)−10=23⋅55(x+7)−10=23⋅5

$\Rightarrow 5\left(x+7\right)-10=40$

$\Rightarrow 5\left(x+7\right)=40+10$

⇒�+7=505⇒x+7=550​

$\Rightarrow x+7=10$

$\Rightarrow x=10-7$

$\Rightarrow x=3$

b) 9�−2⋅32=349x−2⋅32=34

$\Rightarrow 9x-18=81$

$\Rightarrow 9x=81+18$

$\Rightarrow 9x=99$

⇒�=999⇒x=999​

$\Rightarrow x=11$

c) 525⋅5�−1=525525⋅5x−1=5