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

\(a,2^x.4=128\\2^x.2^2=2^7\\ 2^x=\dfrac{2^7}{2^2}=2^{7-2}=2^5\\ Vậy:x=5\\ ----\\ b,\left(2x+1\right)^3=125=5^3\\ \Rightarrow 2x+1=5\\ 2x=5-1=4\\ x=\dfrac{4}{2}=2\\ ----\\ c,2x-2^6=6\\ 2x=6+2^6=6+64\\ 2x=70\\ x=\dfrac{70}{2}=35\\ ----\\ d,49.7^x=2401\\ 7^x=\dfrac{2401}{49}=49=7^2\\ Vậy:x=2\)

a) 128 - 3(x + 4) = 23

             3(x + 4) = 128 -23

             3(x + 4) = 105

                x + 4  = 105 : 3

                x + 4  = 35

                x        = 35 - 4

                x        = 31

\(a.128-3\left(x+4\right)=23.\)

\(3\left(x+4\right)=128-23\)

\(3\left(x+4\right)=105\)

\(x+4=105:3=35\)

\(x=35-4=31\)

\(b.\left[\left(4x+28\right)\cdot3+55\right]:5=35\)

\(\left(4x+28\right)\cdot3+55=35\cdot5=175\)

\(\left(4x+28\right)\cdot3=175-55=120\)

\(4x+28=120:3=40\)

\(4x=40-28=12\)

\(x=12:4=3\)

a, \(128-3.\left(x+4\right)=23\)

\(\Rightarrow\) \(3.\left(x+4\right)=105\)

\(\Rightarrow\) \(x+4=35\)

\(\Rightarrow\) \(x=31\)

b, \(\left[\left(4x+28\right).3+55\right]:5=35\)

\(\Rightarrow\) \(\left(4x+28\right).3+55=175\)

\(\Rightarrow\) \(\left(4x+28\right).3=120\)

\(\Rightarrow\) \(4x+28=40\)

\(\Rightarrow\) \(4x=12\)

\(\Rightarrow\) \(x=3\)

3:

a: 3^x*3=243

=>3^x=81

=>x=4

b; 2^x*16^2=1024

=>2^x=4

=>x=2

c: 64*4^x=16^8

=>4^x=4^16/4^3=4^13

=>x=13

d: 2^x=16

=>2^x=2^4

=>x=4

a) \(2^x\cdot4=128\)

\(2^x=32\)

\(x=5\)

b) \(x^{15}=x\)

\(\Leftrightarrow x\in\left\{0;1\right\}\)

c) \(\left(2x+1\right)^3=125\)

\(\Rightarrow2x+1=5\)

\(2x=4\)

\(x=2\)

a)x=5

b)x=0 hoặc x=1

c)x=2

d)x=5 hoặc x=6

a, x= 5  
b, x= 1
c, x = 2
d, x =6

a)  \(|2x-2|+|3-3x|=125\left(1\right)\)

Ta có: 

\(2x-2=0\Leftrightarrow x=1\)

\(3-3x=0\Leftrightarrow x=1\)

Lập bảng xét dấu :

 

Với \(x< 1\Rightarrow\hept{\begin{cases}2x-2< 0\\3-3x>0\end{cases}\Rightarrow\hept{\begin{cases}|2x-2|=2-2x\\|3-3x|=3-3x\end{cases}}\left(2\right)}\)

Thay (2) vào (1) ta được :

\(\left(2-2x\right)+\left(3-3x\right)=125\)

\(2-2x+3-3x=125\)

\(-5x+5=125\)

\(-5x=120\)

\(x=-24\)( chọn )

Với \(x\ge1\Rightarrow\hept{\begin{cases}2x-2>0\\3-3x< 0\end{cases}}\Rightarrow\hept{\begin{cases}|2x-2|=2x-2\\|3-3x|=3x-3\end{cases}\left(3\right)}\)

Thay (3) vào (1) ta được :

\(\left(2x-2\right)+\left(3x-3\right)=125\)

\(2x-2+3x-3=125\)

\(5x-5=125\)

\(5x=130\)

\(x=26\)9 (CHọn )

Vậy \(x\in\left\{-24;26\right\}\)

b) \(|x-2018|+|x-2019|=1\left(1\right)\)

Ta có: \(x-2018=0\Leftrightarrow x=2018\)

          \(x-2019=0\Leftrightarrow x=2019\)

Lập bảng xét dấu :

+) Với \(x< 2018\Rightarrow\hept{\begin{cases}x-2018< 0\\x-2019< 0\end{cases}\Rightarrow\hept{\begin{cases}|x-2018|=2018-x\\|x-2019|=2019-x\end{cases}\left(2\right)}}\)

Thay (2) vào (1) ta được :

\(\left(2018-x\right)+\left(2019-x\right)=1\)

\(2018-x+2019-x=1\)

\(4037-2x=1\)

\(2x=4036\)

\(x=2018\)( Loại  )

+) Với \(2018\le x< 2019\Rightarrow\hept{\begin{cases}x-2018>0\\x-2019< 0\end{cases}\Rightarrow\hept{\begin{cases}|x-2018|=x-2018\\|x-2019|=2019-x\end{cases}\left(3\right)}}\)

Thay (3) vào (1) ta được :

\(\left(x-2018\right)+\left(2019-x\right)=1\)

\(x-2018+2019-x=1\)

\(1=1\)( luôn đúng )

+) Với \(x\ge2019\Rightarrow\hept{\begin{cases}x-2018>0\\x-2019>0\end{cases}\Rightarrow\hept{\begin{cases}|x-2018|=x-2018\\|x-2019|=x-2019\end{cases}\left(4\right)}}\)

Thay (4) vào (1) ta được :

\(\left(x-2018\right)+\left(x-2019\right)=1\)

\(2x-4037=1\)

\(x=2019\)( Chọn )

Vậy \(2018\le x\le2019\)

a) \(\left(2x+1\right)^3=27\)

\(\Leftrightarrow2x+1=3\)

\(\Leftrightarrow x=1\)

b) \(\left(2x-1\right)^3=125\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow x=3\)

c) \(\left(x+1\right)^4=\left(2x\right)^4\)

\(\Leftrightarrow x+1=2x\)

\(\Leftrightarrow x=1\)

d) \(\left(2x-1\right)^5=x^5\)

\(\Leftrightarrow2x-1=x\)

\(\Leftrightarrow x=1\)

a. ( 2x + 1 )³ = 27

<=> ( 2x + 1 )³ = 3³

<=> 2x + 1 = 3

<=> 2x = 2

<=> x = 1

b. ( 2x - 1 )³ = 125

<=> ( 2x - 1 )³ = 5³

<=> 2x - 1 = 5

<=> 2x = 6

<=> x = 3

c. ( x + 1 )⁴ = 2x⁴

<=> x + 1 = 2x

<=> x = 1

d. ( 2x - 1 )⁵ = x⁵

<=> 2x - 1 = x

<=> x = 1

a) \(\left(2x+1\right)^3=27\)

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

\(\Rightarrow2x+1=3\)

\(2x=3-1\)

\(2x=2\)

\(x=2\div2\)

\(x=1\)

Vậy \(x=1\).

b) \(\left(2x-1\right)^3=125\)

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

\(\Rightarrow2x-1=5\)

\(2x=5+1\)

\(2x=6\)

\(x=6\div2\)

\(x=3\)

Vậy \(x=3\).

c) \(\left(x+1\right)^4=\left(2x\right)^4\)

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

\(x-2x=-1\)

\(-x=-1\)

\(x=1\)

Vậy \(x=1\).

d) \(\left(2x-1\right)^5=x^5\)

\(\Rightarrow2x-1=x\)

\(2x-x=1\)

\(x=1\)

Vậy \(x=1\).

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