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.

a) \(3\left(2x-5\right)+125=134\)

\(\Leftrightarrow3\left(2x-5\right)=9\)

\(\Leftrightarrow2x-5=3\)

\(\Leftrightarrow2x=8\Leftrightarrow x=4\)

b) \(\left(2x+5\right)+\left(2x+3\right)+\left(2x+1\right)=27\)

\(\Leftrightarrow6x+9=27\)

\(\Leftrightarrow6x=18\Leftrightarrow x=3\)

d) \(27\left(x-27\right)-27=0\)

\(\Leftrightarrow27\left(x-27\right)=27\)

\(\Leftrightarrow x-27=1\Leftrightarrow x=28\)

3.(x+5)-2x=27
3.x+3.5-2x=27
3x+15-2x=27
x.(3-2)+15=27
x.  1    +15=27
  x       +15=27
  x             =27-15
  x             =12

 \(\frac{2x-1}{3}=\frac{27}{2x-1}\)

\(\Rightarrow\left(2x-1\right)\left(2x-1\right)=27\cdot3\)

\(\Rightarrow\left(2x-1\right)^2=81\)

\(\Rightarrow\left(2x-1\right)^2=\left(\pm9\right)^2\)

\(\Rightarrow\orbr{\begin{cases}2x-1=9\\2x-1=-9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=10\\2x=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)

ko bt đề đúng ý bn chưa ?

\(\frac{2x-1}{3}=\frac{27}{2x-1}\)

\(\Leftrightarrow\left(2x-1\right)^2=27.3=81\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=9\\2x-1=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=10\\2x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=-4\end{cases}}}\)

Mơn nhiều ạ...

\(a,\left(1-2x\right)^3=-8\)

\(\Rightarrow1-2x=-2\)

\(\Rightarrow2x=3\)

\(\Rightarrow=\frac{3}{2}\)

\(b,\left(2x-1\right)^3=-27\)

\(\Rightarrow2x-1=-3\)

\(\Rightarrow2x=-2\)

\(\Rightarrow x=-1\)

x^3 = 27

x^3 = 3^3

=> x = 3

(2x-1)^3 = 8

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

2x-1 = 2

2x = 2+1

2x = 3

x = 3:2

=> ko có x phù hợp.

(x-2)^2 = 16

(x-2)^2 = 4^2

x-2 = 4

x = 4+2

x = 6

Chúc bn học tốt!

x^3 = 27

x^3 = 3^3

Vậy x = 3

Đề bài 2 hình như sai bạn ạ

( x - 2 )^2 = 16

( x- 2 )^2 = 4^2

x - 2 = 4

x      = 4 + 2

x      = 6

Vậy x = 6

1) \(x^3=27=3^3\Rightarrow x=3\)

2) \(\left(2x-1\right)^3=8=2^3\Rightarrow2x-1=2\)

\(\Rightarrow2x=3\Rightarrow x=\dfrac{3}{2}\)

3) \(\left(x-2\right)^2=16\)

Mà \(x\in N\)

\(\Rightarrow x-2=4\Rightarrow x=6\)

1) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

\(\Leftrightarrow\left(2x-5\right).-2=0\)

\(\Leftrightarrow-4x+10=0\)

\(\Leftrightarrow-4x=-10\)

\(\Leftrightarrow x=\frac{5}{2}.\)

Vậy \(S=\left\{\frac{5}{2}\right\}\)

2)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right).\left(x^2-3x+9+x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)

\(\Leftrightarrow\left(x+3\right).x.\left(x-2\right)=0\)

\(\Leftrightarrow x+3=0\)hoặc \(x=0\)hoặc \(x-2=0\)

\(\Leftrightarrow x=-3\)hoặc \(x=0\)hoặc \(x=2\)

Vậy \(S=\left\{-3;0;2\right\}\)