a) 2^x . 4 = 128

<=> 2^x = 32 

<=> 2^x = 2⁵

<=> x = 5

b) x¹⁵ = x¹

<=> x¹⁵ - x = 0

<=> x(x¹⁴ - 1) = 0

<=> \(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}=1^{14}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

c) (2x + 1)³ = 125

<=> (2x + 1)³ = 5³

<=> 2x + 1 = 5

<=> 2x = 4

<=> x = 2

d) (x - 5)⁴ = (x - 5)⁶

<=> (x - 5)⁶ - (x - 5)⁴ = 0

<=> (x - 5)⁴[(x - 5)² - 1] = 0

<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)

Khi (x - 5)⁴ = 0 => x - 5 = 0 => x = 5

Khi (x - 5)² - 1 = 0 <=> (x - 5)² = 1² <=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

a, 2x . 4 = 128

=> 2x = 128 : 4 = 32

=> x = 32 : 2 = 16

Vậy x = 16

b, x15 = x 1 => Sai đề

c, (2x + 1)³ = 125

=> ( 2x + 1 ) = 5³

=> 2x + 1 = 5 

=> 2x = 5 - 1

=> 2x = 4

=> x = 4 : 2

=> x = 2

a) \(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)

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

\(\Rightarrow2x+1=5\Rightarrow x=2\)

c) \(\Rightarrow2^x=32\Rightarrow x=5\)

d) \(\Rightarrow4^3.4^x=4^5\Rightarrow4^x=4^2\Rightarrow x=2\)

e) \(\Rightarrow3^3.3^x=3^5\Rightarrow3^x=3^2\Rightarrow x=2\)

f) \(\Rightarrow7^2.7^x=7^4\Rightarrow7^x=7^2\Rightarrow x=2\)

a. 2^x . 4 = 128

<=> 2^{x + 2} = 2⁷

<=> x + 2 = 7

<=> x = 5

b. (2x + 1)³ = 125

<=> (2x + 1)³ - 5³ = 0

<=> (2x + 1 - 5)\(\left[\left(2x+1\right)^2+\left(2x+1\right).5+25\right]=0\)

<=> (2x - 4)(4x² + 4x + 1 + 10x + 5 + 25) = 0

<=> (2x - 4)(4x² + 14x + 31) = 0

<=> \(\left[{}\begin{matrix}2x-4=0\\4x^2+14x+31=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\VôNghiệm\end{matrix}\right.\)

c. 2^x - 26 = 6

<=> 2^x = 32

<=> x = 5

d. 64 . 4^x = 4⁵

<=> 4³ . 4^x = 4⁵

<=> 4^{3 + x} = 4⁵

<=> 3 + x = 5

<=> x = 2

e. 27 . 3^x = 243

<=> 3³ . 3^x = 3⁵

<=> 3^{3 + x} = 3⁵

<=> 3 + x = 5

<=> x = 2

g. 49 . 7^x = 2401 (Bn xem lại đề câu này)

<=> 7² . 7^x = 7⁴

<=> 7^{2 + x} = 7⁴

<=> 2 + x = 4

<=> x = 2

a, 2^x.4= 128

=> 2^x= 128:4

=>2^x= 32

=> 2^x=2⁵

=> x=5

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

\(\Rightarrow\left(2x+1\right)^3=\left(\pm5\right)^3\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c, 2^x - 26 = 6

=> 2^x = 32

=> 2^x= 2⁵

=> x=5 

d, 64. 4^x = 4⁵

=> 4³. 4^x= 4⁵

=> 3+x=5

=> x=2

e, 27. 3^x= 243

=> 3³. 3^x= 3⁵

=> 3+x=5

=> x=2

Hoctot

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.`

\(#48Cd\)

Ta có : (x - 5)⁴ = (x - 5)⁶

=>  (x - 5)⁴ - (x - 5)⁶ = 0

=> (x - 5)⁴ (1 - (x - 5)²) = 0

\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)=0\\\left(x-5\right)^2=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x-5=1;-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=4;6\end{cases}}\)

Vậy x = {4;5;6}

Tìm x thuộc N:

2x.4=128

2x    =128:4

2x     =32

 x       =32:2

 x       =16

b.x.15=x

   x thỏa mãn điều kiện:0,1

c.(2x+1)³=125

   (2x+1)³=5³

==>2x+1=5

      2x    =5-1

      2x     =4

        x      =4:2=2

phần d mk chưa hiểu lắm

+) Cấm ghi "nhanh lên nhé ai giải được thì tk cho"

a) 2x.4 = 128

    2x    = 128 : 4

    2x    = 32

    2x    = 2⁵

b) x15 = x

    => x = 0

c) (2x + 1)³ = 125

   (2x + 1)³ = 5³

=> 2x + 1 = 5

=> 2x = 5 - 1

=> 2x = 4

=> x = 2

d) (x - 5).4 = (x - 5).6

=> x = 0

a.2x-5=41-18

 2x-5=23

2x=23+5

2x=28

x=14

a. 41 – (2x – 5) = 18

b. 2x. 4 = 128

Mọi người giúp em nha

a.  41-(2x-5)=18

           2x-5=41-18

            2x-5=23

              2x=23+5

              2x=28

              x=28:2

               x=14

\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)

Câu a) xem lại đề giùm nhé em

b) \(\left(x-1\right)^3=9^3\)

\(x-1=9\)

\(x=10\)

Vậy \(x=10\)

c) \(\left(x-1\right)^2=25\)

\(x-1=5\) hoặc \(x-1=-5\)

* \(x-1=5\)

\(x=6\)

* \(x-1=-5\)

\(x=-4\)

Vậy \(x=-4\); \(x=6\)

d) \(\left(2x+1\right)^3=125\)

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

\(2x+1=5\)

\(2x=4\)

\(x=2\)

Vậy \(x=2\)

e) Sửa đề: \(\left(2x+4\right)^3=64\)

\(\left(2x+4\right)^3=4^3\)

\(2x+4=4\)

\(2x=0\)

\(x=0\)

Vậy \(x=0\)

a) 17-(2x-11) = 12-3x

2x-11 = 17-12+3x

2x-11 = 5+3x

2x-3x = 5+11

-x = 16

x = -16

b, |2x-3|=7

=>2x-3=7 hoặc 2x-3=-7

=>2x=10 hoặc 2x=-4

=>x=5 hoặc x=-2

c, (2x-1)⁴ = 16

(2x-1)⁴ = 2⁴

2x-1=2

2x=3

x=3/2

d, (17-2x)³ = -125

(17-2x)³ = (-5)³

17-2x=-5

2x=17-(-5)

2x=22

x=11

a, => 17-2x+11 = 12-3x

=> 28-2x=12-3x

=> 28=12-3x+2x = 12-x

=> x=12-28 = -16

b, => 2x-3=7 hoặc 2x-3=-7

=> x=5 hoặc x=-2

c, => 2x-1=2 hoặc 2x-1=-2

=> x=3/2 hoặc x=-1.2

d, => 17-2x=-5

=> 2x=17-(-5) = 22

=> x=22:2 = 11

Tk mk nha

a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)

c: Ta có: \(\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b. 

PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$

$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$

$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$

$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$

$\Leftrightarrow (x-3)^2(2x+1)^2=0$

$\Leftrightarrow (x-3)(2x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $2x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$

d.

$x^2-2x=24$

$\Leftrightarrow x^2-2x-24=0$

$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$

$\Leftrightarrow x=-4$ hoặc $x=6$