a.\(\dfrac{1}{3}\) + x  = \(\dfrac{5}{6}\)

       x = \(\dfrac{5}{6}\) - \(\dfrac{1}{3}\)

      x = \(\dfrac{1}{2}\)

b. | x-1| - \(\dfrac{2}{5}\) = \(\dfrac{11}{10}\) 

   | x-1|        = \(\dfrac{11}{10}\) + \(\dfrac{2}{5}\)

  |x-1|        = \(\dfrac{3}{2}\)

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

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

\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

c, \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = 1

            \(\dfrac{2}{3}\) (\(\dfrac{x}{2}\) + 3) = 1 - \(\dfrac{1}{3}\)

             \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = \(\dfrac{2}{3}\)

                   \(\dfrac{x}{2}\) + 3 = 1

                   \(\dfrac{x}{2}\)       = 1 - 3

                    \(\dfrac{x}{2}\)    = -2

                     \(x\) = -4

d, \(\dfrac{x+2}{3}\) = \(\dfrac{27}{x+2}\)

(x+2)² = 27.3

(x+2) =9²

\(\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)

a. (x - 2)² = 1

<=> (x - 2)² = 1² = (-1)²

<=> \(\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\begin{cases}x=3\\x=1\end{cases}\)

Vậy x \(\in\){1; 3}.

b. (2x - 1)³ = -8

<=> (2x - 1)³ = (-2)³

<=> 2x - 1 = -2

<=> 2x = -2 + 1

<=> 2x = -1

<=> x = -1/2

Vậy x = -1/2.

c. (x + 1/2)² = 1/16

<=> (x + 1/2)² = (1/4)² = (-1/4)²

<=> \(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)

Vậy x \(\in\){-1/4; -3/4}.

d. (x - 2)³ = -27

<=> (x - 2)³ = (-3)³

<=> x - 2 = -3

<=> x = -3 + 2

<=> x = -1

Vậy x = -1.

a.\(\left(x-2\right)^2\)=1

<=> x-2=1 hoặc x-2=-1

<=> x= 3 hoặc x=1

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

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

2x-1=-2

2x=-1

x=-1/2

c.\(\left(x+\frac{1}{2}\right)^2\)=\(\frac{1}{16}\)

\(\left(x+\frac{1}{2}\right)^2\)=\(\left(\frac{1}{4}\right)^2\)hoặc \(\left(x+\frac{1}{2}\right)^2\)=\(\left(-\frac{1}{4}\right)^2\)

x+\(\frac{1}{2}\)=\(\frac{1}{4}\)  hoặc x+\(\frac{1}{2}\)=-\(\frac{1}{4}\)

x=-\(\frac{1}{4}\)hoặc x=-\(\frac{3}{4}\)

d.\(\left(x-2\right)^3\)=-27

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

x-2=-3

x=-1

a. x=5/6

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

3^ x -1 = 1/243

3^x =1/243 +1

3^x = 244 / 243

Ta  thấy đây ko phải lũy thừa của 3 => Ko có x thỏa mãn

81^-2x . 27^x =9^5

81^-2 . 81^x . 27^x =9^5

1/9^4 . (81.27)^x =9 ^5

            3^6x = 9^5 : 1/9^4

             3^6x = 9^9

            3^6x = 3^18

    => 6x =18

  x=3

2^x +2^x +3 =144

2.(2^x) =141

2^x+1 = 141

Ta thấy 141 ko phải lũy thừa của 2 => ko có x thỏa mãn

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-8\end{matrix}\right.\)

a) Ta có: \(\left(2x-3\right)-\left(x-5\right)=\left(x+2\right)-\left(x-1\right)\)

\(\Leftrightarrow2x-3-x+5=x+2-x+1\)

\(\Leftrightarrow x+2=3\)

hay x=1

Vậy: x=1

b) Ta có: \(2\left(x-1\right)-5\left(x+2\right)=-10\)

\(\Leftrightarrow2x-2-5x-10=-10\)

\(\Leftrightarrow-3x=-10+10+2=2\)

hay \(x=-\dfrac{2}{3}\)

Vậy: \(x=-\dfrac{2}{3}\)

a, (2x - 3) - (x - 5) = (x + 2) - (x - 1)

 2x - 3 - x + 5 = x + 2 - x + 1

(2x - x) + (-3 + 5) = (x - x) + (2 + 1)

x + 2 = 3

x = 1

a) (X - 2)\(^2\) = 1 <=> X - 2 = \(\sqrt{1}\) <=> X = 1 + 2 <=> X = 3

Bài 1 :

\(C=\frac{1}{\left|x-2\right|+3}\)

\(C\le\frac{1}{3}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy....

Bài 2 :

a) \(\left(\frac{1}{2}\right)^{3x-1}=\frac{1}{32}\)

\(\left(\frac{1}{2}\right)^{3x-1}=\left(\frac{1}{2}\right)^5\)

\(\Rightarrow3x-1=5\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

b) \(2\cdot3^{x-405}=3^{x-1}\)

\(2=3^{x-1}:3^{x-405}\)

\(2=3^{x-1-x+405}\)

\(2=3^{404}\)( vô lí )

=> x thuộc rỗng

c) \(\frac{1}{81}\cdot27^{2x}=\left(-9\right)^4\)

\(\frac{27^{2x}}{81}=9^4\)

\(\frac{\left(3^3\right)^{2x}}{3^4}=\left(3^2\right)^4\)

\(\frac{3^{6x}}{3^4}=3^8\)

\(3^{6x-4}=3^8\)

\(\Rightarrow6x-4=8\)

\(\Rightarrow6x=12\)

\(\Rightarrow x=2\)

d) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

\(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)

\(\left(4x-1\right)^{20}\cdot\left[\left(4x-1\right)^{10}-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}4x-1=0\\4x-1=\left\{\pm1\right\}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=\left\{\frac{1}{2};0\right\}\end{cases}}\)