a) \(\left(2x+1\right)^3=125\)

\(\left(2x+1\right)^3=5^3\)

\(2x+1=5\)

\(2x=4\)

\(x=2\)

b) \(3^x+25=26\times2^2+2\times3^0\)

\(3^x+25=26\times4+2\times1\)

\(3^x+25=106\)

\(3^x=106-25\)

\(3^x=81\)

\(3^x=3^4\)

\(x=4\)

(2x+1)3 = 125 

a)<=> (2x+1)3 = 53

<=> 2x+1 = 5

<=> 2x = 4

<=> x = 2

3^x+25=26 . 2^2 + 2. 3^0

b)3^x+25=104 +2

3^x+25=106

3^x=106+25

3^x=81=3^4

=> x=4                                                                      

a) \(2^x\cdot4=128\)

\(\Rightarrow2^x\cdot2^2=2^7\)

\(\Rightarrow2^{x+2}=2^7\)

\(\Rightarrow x+2=7\)

\(\Rightarrow x=5\)

b) \(\left(2x+1\right)^3=125\)

\(\Rightarrow\left(2x+1\right)^3=5^3\)

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=4:2\)

\(\Rightarrow x=2\)

c) \(2x-2^6=6\)

\(\Rightarrow2x-64=6\)

\(\Rightarrow2x=70\)

\(\Rightarrow x=70:2\)

\(\Rightarrow x=35\)

d) \(64\cdot4^x=45\)

\(\Rightarrow4^3\cdot4^x=45\)

\(\Rightarrow4^{x+3}=45\)

Xem lại đề

e) \(27\cdot3^x=243\)

\(\Rightarrow3^3\cdot3^x=3^5\)

\(\Rightarrow3^{x+3}=3^5\)

\(\Rightarrow x+3=5\)

\(\Rightarrow x=2\)

g) \(49\cdot7^x=2401\)

\(\Rightarrow7^2\cdot7^x=7^4\)

\(\Rightarrow7^{x+2}=7^4\)

\(\Rightarrow x+2=4\)

\(\Rightarrow x=2\)

h) \(3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

k) \(3^4\cdot3^x=3^7\)

\(\Rightarrow3^{x+4}=3^7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

n) \(3^x+25=26\cdot2^2+2\cdot3^0\)

\(\Rightarrow3^x+25=104+2\)

\(\Rightarrow3^x+25=106\)

\(\Rightarrow3^x=81\)

\(\Rightarrow3^x=3^4\)

\(x=4\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`2^x*4 = 128`

`=> 2^x = 128 \div 4`

`=> 2^x = 2^7 \div 2^2`

`=> 2^x = 2^5`

`=> x = 5`

Vậy, `x = 5.`

`b)`

\(\left(2x+1\right)^3=125\)

`=> (2x + 1)^3 = 5^3`

`=> 2x + 1 = 5`

`=> 2x = 5-1`

`=> 2x = 4`

`=> x = 4 \div 2`

`=> x = 2`

Vậy, `x = 2`

`c)`

\(2x-2^6=6\)

`=> 2x = 6+2^6`

`=> 2x = 70`

`=> x = 70 \div 2`

`=> x = 35`

Vậy, `x = 35`

`d)`

\(64\cdot4^x=45\) Bạn xem lại đề

`e)`

`27*3^x = 243`

`=> 3^3 * 3^x = 3^5`

`=> 3^(3 + x) = 3^5`

`=> 3 + x = 5`

`=> x = 5 - 3`

`=> x = 2`

Vậy, `x = 2`

`g)`

`49* 7^x = 2401`

`=> 7^2 * 7^x = 7^4`

`=> 7^(2 + x) = 7^4`

`=> 2 + x = 4`

`=> x = 4 - 2`

`=> x = 2`

Vậy, `x = 2`

`h)`

`3^x = 81`

`=> 3^x = 3^4`

`=> x = 4`

Vậy, `x = 4`

`k)`

`3^4 * 3^x = 3^7`

`=> 3^(4 + x) = 3^7`

`=> 4 + x = 7`

`=> x = 7 - 4`

`=> x = 3`

Vậy, `x = 3`

`n)`

`3^x + 25 = 26*2^2 + 2*3^0`

`=> 3^x + 25 = 104 + 2`

`=> 3^x + 25 = 106`

`=> 3^x = 106 - 25`

`=> 3^x = 81`

`=> 3^x = 3^4`

`=> x = 4`

Vậy, `x = 4.`

\(#48Cd\)

\(\left(7x-11\right)^3=2^5.5^2+2.10^2\)

\(\Rightarrow\left(7x-11\right)^3=32.25+2.100\)

\(\Rightarrow\left(7x-11\right)^3=800+200\)

\(\Rightarrow\left(7x-11\right)^3=1000\)

\(\Rightarrow\left(7x-11\right)^3=10^3\)

\(\Rightarrow7x-11=10\)

\(\Rightarrow7x=21\)

\(\Rightarrow x=3\)

(7\(x\) - 11)³ = 2⁵.5² + 2.10²

(7\(x\) - 11)³ = 1000

(7\(x\) - 11)³ = 10³

7\(x\) - 11 = 10

7\(x\)        = 10 + 11

7\(x\)        = 21

   \(x\)       = 21:7

   \(x\)        = 3

\(\left(7x-11\right)^3=2^5.5^2+2.10^2\)

\(\left(7x-11\right)^3=2^5.5^2+2.\left(2.5\right)^2\)

\(\left(7x-11\right)^3=2^5.5^2+2.2^2.5^2\)

\(\left(7x-11\right)^3=2^5.5^2+2^3.5^2\)

\(\left(7x-11\right)^3=5^2\left(2^5+2^3\right)\)

\(\left(7x-11\right)^3=5^2\left(32+8\right)\)

\(\left(7x-11\right)^3=5^2.40\)

\(\left(7x-11\right)^3=25.40\)

\(\left(7x-11\right)^3=1000\)

\(\left(7x-11\right)^3=10^3\)

\(\Rightarrow7x-11=10\)

\(\Rightarrow7x=10+11\)

\(\Rightarrow7x=21\)

\(\Rightarrow x=21:7\)

\(\Rightarrow x=3\)

\(\left(7x-11\right)^3=2^5\cdot5^2+200\cdot2023^0\\ \left(7x-11\right)^3=32\cdot25+200\cdot1\\ \left(7x-11\right)^3=1000\\ \left(7x-11\right)^3=10^3\\ 7x-11=10\\ 7x=21\\ x=\dfrac{21}{7}=3\)

Lời giải:

$(7x-11)^3=2^5.5^2+200.2023^0=32.25+200=1000=10^3$

$\Rightarrow 7x-11=10$

$\Rightarrow 7x=21$

$\Rightarrow x=3$

(7x - 11)³ = 2⁵.5² + 200

<=> (7x - 11)³  = 32.25 + 200

<=> (7x - 11)³ = 800 + 200

<=> (7x - 11)³ = 1000

<=> (7x - 11)³ = 10³

<=> 7x - 11 = 10

<=> 7x = 21

<=> x = 3 

X=0 

Vì 2 mũ 5 = 32

2 mũ 2 = 4

Suy ra 2 mũ 5 + 2 mũ 2 +200 = 32+4+200=236

* ước của 236 = (1;2;4;59;118;236)

* Mà chỉ có 59+11=70 có chữ số đầu tiên là 7 

ĐÚNG THÌ TICK CHO MÌNH NHA

1: \(6x^2-8=40\)

\(\Leftrightarrow x^2=8\)

hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)

`a) 27.3^x =243`

`3^x=243/27=9`

`3^x=3^2`

`=>x=2`

`b) 49.7^x =2401`

`7^x = 2401/49=49`

`7^x=7^2`

`=>x=2`

`c) 3^x +25 =26*2^2+2.3^0`

`3^x = 26.4+2.1-25 = 81`

`3^x =3^4`

`=>x=4`

\(e,27.3^x=243\\ \Leftrightarrow3^3.3^x=3^5\\ \Leftrightarrow3^{3+x}=3^5\\ \Rightarrow3+x=5\\ \Leftrightarrow x=2\\ g,49.7^x=2401\\ \Leftrightarrow7^2.7^x=7^4\\ \Leftrightarrow7^{2+x}=7^4\\ \Rightarrow2+x=4\\ \Leftrightarrow x=2\\ h,3^x+25=26.2^2+2.3^0\\ \Leftrightarrow3^x+25=26.4+2.1\\ \Leftrightarrow3^x=104+2-25\\ \Leftrightarrow3^x=81=3^4\\ Vậy:x=4\)

x=17.17992192

3^x + 25 = 26 . 2² + 2 . 30

=> 3^x + 25 = 26 . 4 + 60

=> 3^x + 25 = 164

=> 3^x = 164 - 25

=> 3^x = 139

=> Không tồn tại bất cứ giá trị x nào thỏa mãn

(Đề có sai không?)