a) x = 2 

b) x = 2     

c) x = 2

d) x = 1.

a) Ta có:  ( 2 x + 1 ) 3 = 3 3 nên 2x + 1 = 3. Do đó x = 1.

b) Ta có: ( 2 x - 1 ) 3 = 5 3 nên 2x - 1 = 5. Do đó x = 3.

( 2x + 1 ) ³ = 125

( 2x + 1 ) ³ = 5³

2x + 1 = 5

2x = 5 - 1

2x = 4

x = 4 : 2

x = 2

(2x+1)³=125

2x+1=5

2x=5-1

2x=4

x=4:2

x=2

`@` `\text {Ans}`

`\downarrow`

`a)`

\(5\cdot x^3-5=0\)

`=> 5*x^3 = 0+5`

`=> 5*x^3 = 5`

`=> x^3 = 5 \div 5`

`=> x^3 = 1`

`=> x^3 = 1^3`

`=> x=1`

Vậy, `x=1.`

`b)`

\(( x+1)^2 = 16\)

`=> (x+1)^2 = (+-4)^2`

`=>`\(\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4-1\\x=-4-1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy, `x \in {3; -5}`

`c)`

\(( x+1)^3 = 27\)

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

`=> x+1=3`

`=> x=3-1`

`=> x=2`

Vậy, `x=2.`

`d)`

\(( x-1)^3 = 343\)

`=> (x-1)^3 = 7^3`

`=> x-1=7`

`=> x=7+1`

`=> x=8`

Vậy, `x=8.`

`e)`

\((2x - 1^3) = 125\) hay đề là `(2x-1)^3 = 125` vậy ạ?

Mình làm cả 2 TH nhé!

`(2x-1^3)=125`

`=> 2x-1=125`

`=> 2x=125+1`

`=> 2x=126`

`=> x=126 \div 2`

`=> x=63`

TH2:

`(2x-1)^3 = 125`

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

`=> 2x-1=5`

`=> 2x=5+1`

`=> 2x=6`

`=> x=6 \div 2`

`=> x=3`

Vậy, `x=3.`

(a) \(5x^3-5=0\Leftrightarrow5x^3=5\Leftrightarrow x^3=1\Leftrightarrow x=1\)

(b) \(\left(x+1\right)^2=16\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

(c) \(\left(x+1\right)^3=27\Leftrightarrow x+1=3\Leftrightarrow x=2\)

(d) \(\left(x-1\right)^3=343\Leftrightarrow x-1=7\Leftrightarrow x=8\)

(e) \(\left(2x-1\right)^3=125\Leftrightarrow2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)

(2x + 1)³ = 125

=> (2x + 1)³ = 5³

=> 2x + 1 = 5

=> 2x = 5 - 1

=> 2x = 4

=> x = 4 : 2

=> x = 2

Vậy x = 2

( 2x + 1 )³ = 125

( 2x + 1 )³ = 5³

=> 2x + 1 = 5

     2x   = 5 - 1

      2x  =   4

      x = 4 : 2

      x =    2

x=2

(2x + 1)3 = 125

(2x + 1)  = 125 : 3

(2x + 1)  = \(\frac{125}{3}\)

 2x        = \(\frac{125}{3}\)+1

2x        = \(\frac{128}{3}\)

 x         = \(\frac{128}{3}\): 2

 x         = \(\frac{128}{6}\)

(2x + 1)3 = 125

(2x + 1) = 125 : 3

2x + 1 = \(\frac{125}{3}\)

2x = \(\frac{125}{3}-1\)

2x = \(\frac{122}{3}\)

x = \(\frac{122}{3}:2\)

x = \(\frac{61}{3}\)

(2x + 1)3 = 125

2x + 1     = 125 : 3

2x + 1     = 125/3

2x           =  125/3 - 1

2x           = 122/3

x             = 122/3 : 2

x             = 61/3

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

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 \(x=0\)