vì 5^3 = 125 

=> 1/125 - 1/125 = 0 

=> (1/125-1^3)x(1/125-2^3)x...0x...x(1/125-25^3=0

a: x^3=7^3

=>x^3=343

=>\(x=\sqrt[3]{343}=7\)

b: x^3=27

=>x^3=3^3

=>x=3

c: x^3=125

=>x^3=5^3

=>x=5

d: (x+1)^3=125

=>x+1=5

=>x=4

e: (x-2)^3=2^3

=>x-2=2

=>x=4

f: (x-2)^3=8

=>x-2=2

=>x=4

h: (x+2)^2=64

=>x+2=8 hoặc x+2=-8

=>x=6 hoặc x=-10

j: =>x-3=2 hoặc x-3=-2

=>x=1 hoặc x=5

k:

9x^2=36

=>x^2=36/9

=>x^2=4

=>x=2 hoặc x=-2

l:

(x-1)^4=16

=>(x-1)^2=4(nhận) hoặc (x-1)^2=-4(loại)

=>x-1=2 hoặc x-1=-2

=>x=3 hoặc x=-1

1. \(\left(x+1\right)^3-125\)

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

\(=\left(x+1-5\right).\left[\left(x+1\right)^2+\left(x+1\right).5+5^2\right]\)

2. \(\left(x+4\right)^3-64\)

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

\(=\left(x+4-4\right).\left[\left(x+4\right)^2+\left(x+4\right).4+4^2\right]\)

3. \(x^3-\left(y-1\right)^3\)

\(=(x^3-y+1).\left[\left(x^2\right)+x.\left(y+1\right)+\left(y+1\right)^2\right]\)

\(\)4. \(\left(a+b\right)^3-c^3\)

\(=\left[\left(a+b\right)-c\right].\left[\left(a+b\right)^2+\left(a+b\right).c+c^2\right]\)

5. \(125-\left(x+2\right)^3\)

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

\(=\left(5-x-2\right).\left[5^2+5.\left(x+2\right)+\left(x+2\right)^2\right]\)

6. \(\left(x+1\right)^3+\left(x-2\right)^3\)

\(=\left[\left(x+1\right)+\left(x-2\right)\right].\left[\left(x+1\right)^2-\left(x+1\right).\left(x-2\right)+\left(x-2\right)^2\right]\)

\(2^x=32=>x=5\)

\(2^{x+1}=32=>x+1=5=>x=4\)

\(x^3=125=>x=5\)

\(\left(x+1\right)^3=125=>x+1=5=>x=4\)

3. x³ = 125 => x = 5

4. (x + 1)³ = 125 

   (x + 1)³ = 5³

=> x + 1 = 5

=> x = 4 

t i c k nhé!! 45457474678789886973463125135153152454654657657868768

a,       

2^x = 32

2^x = 2⁵

=> x = 5

b,       

2^{x + 1} = 32

2^{x + 1} = 2⁵

=> x + 1 = 5

     x       = 5 - 1

     x       = 6

c,

x³ = 125

x³ = 5³

=> x = 5

d, 

( x - 1 )³ = 125

( x - 1 )³ = 5³

=> x - 1 = 5

     x      = 5 + 1 

     x      = 6

a: \(\Leftrightarrow4^{x-5}\cdot17=68\)

=>4^x-5=4

=>x-5=1

=>x=6

b: \(\Leftrightarrow\dfrac{1}{3}:\left|2x-1\right|=\dfrac{1}{3}+\dfrac{2}{3}=1\)

=>|2x-1|=1/3

=>2x-1=1/3 hoặc 2x-1=-1/3

=>x=2/3 hoặc x=1/3

c: =>|2x-2|=|3x+15|

=>3x+15=2x-2 hoặc 3x+15=-2x+2

=>x=-17 hoặc x=-13/5

a) x = 3

b) x = 2

c) x = 4

d) x = 1.

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

ai giải hộ mình mình k liền

\(\left(2x+1\right)^3=125\\ \Rightarrow\left(2x+1\right)^3=5^3\\ \Rightarrow2x+1=5\\ \Rightarrow2x=4\\ \Rightarrow x=2.\\ b,\left(2x-1\right)^4=16\\ \Rightarrow\left(2x-1\right)^4=2^4\\ \Rightarrow2x-1=2\\ \Rightarrow2x=3\\ \Rightarrow x=\dfrac{3}{2}.\\ c,6.3^x-2.3^x=36\\ \Rightarrow3^x.\left(6-2\right)=36\\ \Rightarrow3^x.4=36\\ \Rightarrow3^x=9\\ \Rightarrow3^x=3^2\\ \Rightarrow x=2.\\ d,2^{x+1}-2^x=32\\ \Rightarrow2^x.\left(2-1\right)=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5.\)