\(=x^3+27-x^3+54=81\)

ta có : 

\(\frac{x+1}{3}+\frac{x+1}{9}+\frac{x+1}{27}+\frac{x+1}{81}=\left(x+1\right)\times\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)\)

\(=\left(x+1\right)\times\frac{40}{81}=\frac{56}{81}\text{ nên }x+1=\frac{56}{40}=\frac{7}{5}\)

vậy \(x=\frac{7}{5}-1=\frac{2}{5}\)

16/81

Có tính k bạn

...

`@` `\text {Ans}`

`\downarrow`

\(\left(-3\right)^x\div81=-27\)

`=>`\(\left(-3\right)^x=-27\cdot81\)

`=>`\(\left(-3\right)^x=\left(-3\right)^3\cdot3^4\)

`=>`\(\left(-3\right)^x=\left(-3\right)^7\)

`=> x=7`

Vậy, `x=7.`

`@` `\text {Kaizuu lv u}`

(-3)^x . 81 = -27

<=> (-3)^x . (-3)⁴ = (-3)³

<=> (-3)^x+4 = (-3)³

=> x+4=3

<=>x= -1

Vũ Hồng Linh bạn check lại bài đầu dùm =_=" 

\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)

\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)

\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)

\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)

\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)

\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)

=> Làm nốt 

Mấy bài kia cũng làm tương tự

(- \(\dfrac{1}{3}\))³.\(x\) = \(\dfrac{1}{81}\)

          \(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))³

          \(x\) =  - (\(\dfrac{1}{3}\))⁴ :(\(\dfrac{1}{3}\))³

           \(x=-\dfrac{1}{3}\)

Vậy \(x=-\dfrac{1}{3}\)

(3\(x\) + 1)³ =  - 27

(3\(x\) + 1)³ = (-3)³

3\(x\) + 1 = -3

 3\(x\)       = - 3 - 1

 3\(x\) = -4

    \(x=-\dfrac{4}{3}\)

Vậy \(x=-\dfrac{4}{3}\)

-3^x/81 = - 27

-3^x = -27 . 81

-3^x = - 2187

\(\Rightarrow\)-3 ^ x = (-3 ) ^ 7

\(\Rightarrow\)x = 7

bằng -27 chứ bạn xem lại đề bài đi

Ta có: \(\dfrac{\left(-3\right)^x}{81}=27\)

\(\Leftrightarrow\left(-3\right)^x=2187\)

hay \(x\in\varnothing\)