a ) 126 - 2x = 12 - ( -28 )

     126 - 2x = 40

     2x = 126 - 40

     2x = 86 

     x = 86 : 2

     x = 43

b ) 25 - 5( x + 1 ) = 30 - 125

     25 - ( 5x + 5 ) = -95

     25 - 5x - 5 = -95

     20 - 5x = -95

     5x = 20 - ( - 95 )

     5x = 115

     x = 115 : 5

     x = 23

c ) I 3x - 1 I - 25 = 73 

     I 3x - 1 I = 73 + 25

     I 3x - 1 I = 98

=> 3x - 1 = 98 hoặc 3x - 1 = -98

=> 3x = 98 + 1 hoặc 3x = -98 + 1

=> 3x = 99 hoặc 3x = -97

=> x = 99 : 3 hoặc x = -97 : 3 

=> x = 33 hoặc x = -93/3

d ) 2( x + 2 ) - 6 = 120 - 4x 

     2x + 4 - 6 = 120 - 4x 

     2x - 2 = 120 - 4x

     2x + 4x = 120 + 2

     6x = 122

     x = 122 : 6

     x = 61/3

a) x = 43

b) x = 23

c) x = 33 hoac x= -97/3

nêu cách làm đi

a) \(\left(2x+1\right)^3=125\)

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

\(2x+1=5\)

\(2x=4\)

\(x=2\)

b) \(3^x+25=26\times2^2+2\times3^0\)

\(3^x+25=26\times4+2\times1\)

\(3^x+25=106\)

\(3^x=106-25\)

\(3^x=81\)

\(3^x=3^4\)

\(x=4\)

(2x+1)3 = 125 

a)<=> (2x+1)3 = 53

<=> 2x+1 = 5

<=> 2x = 4

<=> x = 2

3^x+25=26 . 2^2 + 2. 3^0

b)3^x+25=104 +2

3^x+25=106

3^x=106+25

3^x=81=3^4

=> x=4                                                                      

b) Ta có: \(-5+\left|3x-1\right|+6=\left|-4\right|\)

\(\Leftrightarrow\left|3x+1\right|+1=4\)

\(\Leftrightarrow\left|3x+1\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=3\\3x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{3};-\dfrac{4}{3}\right\}\)

c) Ta có: \(\left(x-1\right)^2=\left(x-1\right)^4\)

\(\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)

\(\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)^2\cdot\left[\left(x-1\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-1-1\right)\left(x-1+1\right)=0\)

\(\Leftrightarrow x\cdot\left(x-1\right)^2\cdot\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{0;1;2\right\}\)

d) Ta có: \(5^{-1}\cdot25^x=125\)

\(\Leftrightarrow5^{-1}\cdot5^{2x}=5^3\)

\(\Leftrightarrow5^{2x-1}=5^3\)

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=4\)

hay x=2

Vậy: x=2

\(\frac{125^{3x}}{25^x}=5^{14}=>\frac{\left(5^3\right)^{3x}}{\left(5^2\right)^x}=5^{14}=>\frac{5^{9x}}{5^{2x}}=5^{14}=>5^{7x}=5^{14}=>7x=14=>x=2\)

đơn giản thôi bạn

3 (x + 25) 

= 3x + 3*25  (bạn nhân 3 với từng số trong ngoặc)           (*là nhân nhé)

= 3x + 75

-125 + 3x = 25 + (-15) - 2x

3x + 2x = 25 + (-15) + (-125)

5x         =    10          + (-125)

5x         =           -115

x           = -115 : 5

x           =     -23

Vậy x = -23

Nguyễn Hoàng Nam chuyển -125 từ vế trái sang vế phải sai rồi đấy.

ồ xin lỗi mk nhầm

( 3x - 1 )³ = 125

( 3x - 1 )³ = 5³

=> 3x - 1 = 5

      3x    = 6

        x    = 2

( 10 + x ) : 25 = 3⁰ . 2²

( 10 + x ) : 25 = 4

10 + x = 100

x = 90