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.

a) x = 2 

b) x = 2     

c) x = 2

d) x = 1.

(2x+1)3=125

=> 2x + 1 = 41,666667

=> 2x = 42,66667

=> x =  21,33333

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

\(2x+1=125.3\)

\(2x+1=375\)

\(2x=375-1\)

\(2x=374\)

\(x=374:2\)

\(x=187\)

`@` `\text {Ans}`

`\downarrow`

`2^x * 4 = 128`

`=> 2^x = 128 * 4`

`=> 2^x = 512`

`=> 2^x = 2^9`

`=> x = 9`

Vậy, `x = 9`

`x^15 = x`

`=> x^15 - x = 0`

`=> x(x^14 - 1) = 0`

`=>` TH1: `x = 0`

`TH2: x^14 - 1 = 0`

`=> x^14 = 1`

`=> x = 1`

Vậy, `x \in {0; 1}`

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

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

`=> 2x + 1 = 5`

`=> 2x = 5 - 1`

`=> 2x =4`

`=> x = 4 \div 2`

`=> x = 2`

Vậy,` x = 2.`

`(x - 5)^4 = (x-5)^6`

`=> (x-5)^4 - (x-5)^6 = 0`

`=> (x-5)^4 * [ 1 - (x-5)^2] = 0`

`=> - (x-6)(x-5)^4(x-4) = 0`

`TH1: (x - 5)^4 = 0`

`=> x - 5 = 0`

`=> x = 0 +5`

`=> x = 5`

`TH2: x - 6=0`

`=> x=6`

`TH3: x-4=0`

`=> x = 4`

Vậy, `x \in {4; 5; 6}`

a: =>2^x=32

=>x=5

b: =>x^15-x=0

=>x(x^14-1)=0

=>x=0; x=1;x=-1

c: =>2x+1=5

=>2x=4

=>x=2

d: =>(x-5)^4[(x-5)^2-1]=0

=>(x-5)(x-4)(x-6)=0

=>x=5;x=4;x=6

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

\(\Leftrightarrow2x+1=15\)

\(\Leftrightarrow2x=14\Leftrightarrow x=7\)

\(\left(2x+1\right)^3=125\) 

\(\left(2x+1\right)^3=5^3\) 

\(2x+1=5\) 

\(x=2\)

b)  2 x - 1 3 = 125

2 x - 1 3  = 5 3

2x – 1 = 5

2x = 5 + 1

2x = 6

x = 6 : 2 = 3

a) \(\left(2x-1\right)^3=27\)

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

\(\Rightarrow2x-1=3\)

\(\Rightarrow2x=3+1=4\)

\(\Rightarrow x=4:2=2\)

b) \(\left(2x+1\right)^3=125\)

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

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=5-1=4\)

\(\Rightarrow x=4:2=2\)

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

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

\(\Rightarrow2=2x-2=1x\)

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

d) \(\left(2x-1\right)^7=x^7\)

\(\Rightarrow2x-1=x\)

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

\(\Rightarrow-1=-x\)

\(\Rightarrow x=1\)

a) x = 2

b) x = 2

c) x = 2

d) x = 1