1/2* x+2/3=9/2

1/2 * x = 9/2 - 2/3 

1/2 * x= 23/6

x= 23/6 : 1/2

x= 23/6 x 2= 23/3

___

1/2*x-1/3=2/3

1/2*x = 2/3 + 1/3

1/2 * x= 1

x= 1: 1/2 

x= 2

____

1/4+3/4:x=3

3/4 : x = 3 - 1/4

3/4 : x= 11/4

x= 11/4 : 3/4

x= 11/3

\(\dfrac{1}{2}\)\(\times\)\(x\) + \(\dfrac{2}{3}\) = \(\dfrac{9}{2}\)

\(\dfrac{1}{2}\)\(\times\)\(x\)        = \(\dfrac{9}{2}\) - \(\dfrac{2}{3}\)

\(\dfrac{1}{2}\)\(\times\)\(x\)       = \(\dfrac{23}{6}\)

      \(x\)       = \(\dfrac{23}{6}\):\(\dfrac{1}{2}\)

      \(x\)      = \(\dfrac{23}{3}\) 

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

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

\(\dfrac{1}{2}\times\)\(x\)      =  1

     \(x\)       = 1 : \(\dfrac{1}{2}\)

   \(x\)         = 2

\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\): \(x\) = 3

          \(\dfrac{3}{4}\): \(x\) = 3 - \(\dfrac{1}{4}\) 

          \(\dfrac{3}{4}\):\(x\) = \(\dfrac{11}{4}\)

              \(x\) = \(\dfrac{3}{4}\): \(\dfrac{11}{4}\)

             \(x\) = \(\dfrac{3}{11}\)

` @ L I N H `

1/2* x+2/3=9/2

1/2 * x = 9/2 - 2/3 

1/2 * x= 23/6

x= 23/6 : 1/2

x= 23/6 x 2= 23/3

___

1/2*x-1/3=2/3

1/2*x = 2/3 + 1/3

1/2 * x= 1

x= 1: 1/2 

x= 2

____

1/4+3/4:x=3

3/4 : x = 3 - 1/4

3/4 : x= 11/4

x= 11/4 : 3/4

x= 11/3

Lời giải:
$x-\frac{x}{3}\times \frac{3}{2}=2-\frac{1}{2}$

$x-x\times \frac{1}{2}=\frac{3}{2}$

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

$x\times \frac{1}{2}=\frac{3}{2}$

$x=\frac{3}{2}: \frac{1}{2}=3$

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

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

\(\dfrac{3x-2}{6}\) - \(\dfrac{4x-6}{6}\) = \(\dfrac{6x}{6}\)

3\(x-2-4x\) + 6 = 6\(x\) 

-\(x\) + 4 - 6\(x\) = 0

7\(x\)  = 4

    \(x\) =  \(\dfrac{4}{7}\) 

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{x-3}{x+1}=\dfrac{x^2}{x^2-1}\)

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

Suy ra: \(x^2-4x+3-x^2=0\)

\(\Leftrightarrow-4x=-3\)

hay \(x=\dfrac{3}{4}\)(thỏa ĐK)

Vậy: \(S=\left\{\dfrac{3}{4}\right\}\)

\(=\left(1-x\right)\left(6a+2a^2\right)=2a\left(3+2a\right)\left(1-x\right)\\ 2,=\left(x-5\right)\left(x-3-2\right)=\left(x-5\right)^2\)

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

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow-x^2+6x-3=-x^2+3x+1\)

\(\Leftrightarrow3x=4\)

hay \(x=\dfrac{4}{3}\)

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

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)

B=2x^2+6x-3x^2-3=x+1-x(x-2)=0

  =-x^2+6x-3=x+1-x^2+x=0

  =4x-3=0

x=3/4

\(\left(-3\sqrt{x}+2\right)\left(x\sqrt{x}+4\sqrt{x}+1\right)\)

\(=-3\sqrt{x}\left(x\sqrt{x}+4\sqrt{x}+1\right)+2\left(x\sqrt{x}+4\sqrt{x}+1\right)\)

\(=-3x^2-12x-3\sqrt{x}+2x\sqrt{x}+8\sqrt{x}+2\)

\(=-3x^2-12x+5\sqrt{x}+2x\sqrt{x}+2\)