a)125 : x = 2² - (-1)

125 : x = 4 + 1

125 : x = 5

x = 125 : 5

x = 25

-------------------------------------------------

b) 2x - 8 = -4

2x = (-4) + 8

2x = 4

x = 4 : 2 

x = 2

-----------------------------------------------------------

c) Xem lại đề.

\(125:x=2^2-\left(-1\right)\)

\(=>125:x=4+1\)

\(=>125:x=5\)

\(=>x=125:5\)

\(=>x=25\)

_____

\(2x-8=-4\)

\(=>2x=\left(-4\right)+8\)

\(=>2x=4\)

\(=>x=4:2\)

\(=>x=2\)

_______

\(6^{2x+5}=216\)

\(=>6^{2x+5}=6^3\)

\(=>2x+5=3\)

\(=>2x=3-5\)

\(=>2x=-2\)

\(=>x=\left(-2\right):2\)

\(=>x=-1\)

\(#NqHahh\)

125:x=2² - (-1)

125:x=4+1

125:x=5

X=125:5

X=25

2x -8 = -4

2x=-4+8

2x=4

X=4:2

X=2

6²(x+5)  = 216

36(x+5) = 216

X+5 = 216:36

X+5=6

X=6-5

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

a) bạn xem thử có số nào mũ 3 lên bằng 981 không nếu ra đọc đi mình giải

b) \(5^x+5^{x+2}=650\)

   \(5^x+5^x+5^2=650\)

   \(5^x\left(1+25\right)=650\)

   \(5^x.26=650\)

  \(5^x=25\)

   \(5^x=5^2\)

Vậy x = 2 

c) \(\left(2x+1\right)^3=125\)

     \(\left(2x+1\right)^3=5^3\)

     \(2x+1=5\)

      \(2x=4\)

         \(x=2\)

Vậy x = 2 

5^x=125

=>5^xx=5³

=> x=3

Vậy x=3

3^2x=81

=>3^2x=3⁴

=>2x=4

=>x=4:2=2

Vậy x=2

Tk mk

a, => 5^x = 5^3

=> x = 3

b, => 3^2x = 3^4

=> 2x = 4

=> x = 2

Tk mk nha

a,5^x=125

5^x=5³

=>x=3

b,3^2x=81

3^2x  = 3⁴

=>. 2x = 4

=> x = 4 : 2

=> x = 2

~~ học tốt ~~

a. 

(2x+1)² = 25 = 5²

=> 2x + 1 = 5

=> 2x = 4

=> x = 2

Vậy x = 2

b.

5^2x+1 = 125 = 5³

=> 2x + 1 = 3

=> 2x = 2

=> x = 1 

Vậy x = 1

Chúc em học tốt!!!

a, ( 2x + 1 )² = 25

<=> ( 2x + 1 )² = 5²

=> 2x + 1 = 5

=> 2x = 4

=> x = 2

b 5^{2x + 1} = 125

<=> 5^2x+1  = 5³

=> 2x + 1 = 3

=> 2x = 2

=> x = 1

\(\left(2x+1\right)^2=25\)

\(\left(2x+1\right)^2=\left(\pm5\right)^2\)

\(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

\(5^{2x+1}=125\)

\(\Rightarrow5^{2x+1}=5^3\)

\(\Rightarrow2x+1=3\)

\(\Rightarrow x=2\)

a) \(\sqrt{2x}=12\left(đk:x\ge0\right)\)

\(2x=144\)

\(x=72\)

b) \(\sqrt{9x^2-6x}+1=10\)\(\left(Đk:x\le0;x\ge\dfrac{2}{3}\right)\)

\(\sqrt{9x^2-6x}=9\)

\(9x^2-6x=81\)

\(\left(3x-1\right)^2=82\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{82}+1}{3}\\x=\dfrac{1-\sqrt{82}}{3}\end{matrix}\right.\)

c) \(x^2\sqrt{5}-\sqrt{125}=0\)

\(x^2\sqrt{5}=5\sqrt{5}\)

\(x^2=5\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)

các thầy cô giúp e vs ạ