\(4^{x-5}=16\)

\(4^{x-5}=4^2\)

\(x-5=2\)

\(x=2+5\)

\(x=7\)

\(45-2^{x-1}=29\)

\(2^{x-1}=16\)

\(2^{x-1}=2^4\)

\(x-1=4\)

\(x=5\)

\(\left(2+x\right)^2=144\)

\(\left(2+x\right)^2=12^2\)

\(2+x=12\)

\(x=12-2\)

\(x=10\)

\(\left(x-5\right)^2=81\)

\(\left(x-5\right)^2=9^2\)

\(x-5=9\)

\(x=14\)

\(\left(13-x\right)^4=81\)

\(\left(13-x\right)^4=3^4\)

\(13-x=3\)

\(x=13-3\)

\(x=10\)

\(...4^{x-5}=4^2\Rightarrow x-5=2\Rightarrow x=7\)

\(...2^{x-1}=45-29=16\Rightarrow2^{x-1}=2^4\Rightarrow x-1=4\Rightarrow x=5\)

\(...\Rightarrow\left(2+x\right)^2=12^2\Rightarrow\left[{}\begin{matrix}2+x=12\\2+x=-12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-14\end{matrix}\right.\)

\(...\Rightarrow\left(x-5\right)^2=9^2\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

\(...\Rightarrow\left(13-x\right)^4=3^4\Rightarrow\left[{}\begin{matrix}13-x=3\\13-x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=16\end{matrix}\right.\)

Mình quên mũ 2 với mũ 4 là chia ra 2 trường hợp, sửa lại nhé:

\(\left(2+x\right)^2=144\)

 \(\left(2+x\right)^2=12^2\)

Th1: 

\(2+x=12\)

\(x=10\)

Th2: 

\(2+x=-12\)

\(x=-14\)

\(\left(x-5\right)^2=81\)

\(\left(x-5\right)^2=9^2\)

Th1: 

\(x-5=9\)

\(x=14\)

Th2: 

\(x-5=-9\)

\(x=-4\)

\(\left(13-x\right)^4=81\)

\(\left(13-x\right)^4=3^4\)

Th1: 

\(13-x=3\)

\(x=10\)

Th2: 

\(13-x=-3\)

\(x=16\)

a: =>1/3:x=3/5-2/3=9/15-10/15=-1/15

=>x=-1/3:1/15=5

b: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)

=>x*2/3=-1

=>x=-3/2

c: =>2x+1=4 hoặc 2x+1=-4

=>x=3/2 hoặc x=-5/2

h: =>x-3=4

=>x=7

g: =>2x-1=3

=>2x=4

=>x=2

f: \(\Leftrightarrow x\cdot\left(\dfrac{3}{2}-\dfrac{7}{3}\right)=\dfrac{3}{2}-\dfrac{2}{3}\)

=>x*-5/6=5/6

=>x=-1

d: =>|2x-1|=3

=>2x-1=3 hoặc 2x-1=-3

=>x=-1 hoặc x=2

(2x-1)⁴ =81=3⁴

^{2x - 1 =3}

x = 4/2 = 2

=>(2x-1)⁴=3⁴

=>2x-1   =3

=>2x     =3+1=4

=>x       = 4:2=2

vậy x = 2 k mk nha

(2x-1)⁴=81

=>(2x-1)⁴=3⁴

=>2x-1=3

=>2x=3+1

=>2x=4

=>x=4:2

=>x=2

\(4.\left(3-2x\right)^2=\frac{1}{81}\)

\(\left(3-2x\right)^2=\frac{1}{324}\)

\(\Leftrightarrow\orbr{\begin{cases}3-2x=\frac{1}{18}\\3-2x=\frac{-1}{18}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{53}{36}\\x=\frac{55}{36}\end{cases}}}\)

Vậy \(x\in\left\{\frac{53}{36};\frac{55}{36}\right\}\)

Cảm ơn nhé!

zko có gì

a) x + 2006 = 2021

x= 2021 - 2006

x= 15

b) 2x - 2016 = 2 ⁴  . 4

2x - 2016 = 64

2x = 64 + 2016

2x = 2080

x= 2080 : 2

x= 1040

c) 3. ( 2x + 1) ³ =81

( 2x-1)³ = 27

( 2x-1)³ = 3³

^{=> 2x-1 = 3}

^{2x= 2}

^{x= 1}

a, \(x\) + 2006 = 2021

    \(x\)             = 2021 - 2006

     \(x\)             = 15

b, 2\(x\) - 2016 = 2⁴.4

    2\(x\) - 2016  = 64

    2\(x\)               = 64 + 2016

    2\(x\)                = 2080

      \(x\)                = 2080 : 2

       \(x\)                = 1040

(2x-1)⁴=81

=> 2x-1= 3 hoặc 2x-1= -3

=> 2x= 3+1 hoặc 2x=-3 +1

=> 2x= 4 hoặc 2x= -2

=> x= 4:2 hoặc x= -2: 2

=> x= 2 hoặc x= -1

(2x-1)⁴=81

=> 2x-1= 3 hoặc 2x-1= -3

=> 2x= 3+1 hoặc 2x=-3 +1

=> 2x= 4 hoặc 2x= -2

=> x= 4:2 hoặc x= -2: 2

=> x= 2 hoặc x= -1

(2x-1)⁴=81

=> 2x-1= 3 hoặc 2x-1= -3

=> 2x= 3+1 hoặc 2x=-3 +1

=> 2x= 4 hoặc 2x= -2

=> x= 4:2 hoặc x= -2: 2

=> x= 2 hoặc x= -1

a) (x - 2)^2 = 16

(x - 2)^2 = 4^2 = (-4)^2

=> x - 2 = 4 hoặc x - 2 = -4

=> x = 4 + 2 = 6

=> x = (-4) + 2 = -2

Vậy x = 6 hoặc x = -2

b) (2x - 1)^3 = 8

(2x - 1)^3 = 2^3

=>2x - 1 = 2

2x = 2 + 1 = 3

\(x=\frac{3}{2}\)

c) (2x - 1)^4 = 81

(2x - 1)^4 = 3^4=(-3)^4

=> 2x - 1 = 3 hoặc 2x - 1 = -3

=>2x = 3 + 1  = 4

=>x = 4 : 2 = 2

2x = (-3) + 1 = -2

x = -1

Vậy x = 2 hoặc x = -1

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