giải các phương trình sau: a,2(3x – 1) – 3x = -10 b,4x^2 – 9=0 c, x + 1/x – 2 – 3 – x/x + 2 = 2x^2 – 4/x^2 – 4 d, /2x – 3/ – 5 = 4x

a,  2(3x-1)-3x=-10

⇔   6x-2-3x  =-10

⇔      3x        =-8

⇔       x         =$\frac{-8}{3}$ 

     Vậy S={$\frac{-8}{3}$} 

b,   4$x^{2}$-9 =0 

⇔ (2x-3)(2x+3)=0

⇔ \(\left[ \begin{array}{l}2x-3=0\\2x+3=0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x= \frac{3}{2} \\x= \frac{-3}{2} \end{array} \right.\)

     Vậy S={±$\frac{3}{2}$}

c, ĐKXĐ: x$\neq$ ±2

    $\frac{x+1}{x-2}$-$\frac{3-x}{x+2}$=$\frac{2x^{2}-4}{x^{2}-4 }$

⇔ $\frac{(x+1)(x+2)}{(x-2)(x+2)}$- $\frac{(3-x)(x-2)}{(x-2)(x+2)}$=$\frac{2x^{2}-4}{(x-2)(x+2) }$

⇒ $x^{2}$+3x+2-3x+6+$x^{2}$-2x= 2$x^{2}$-4

⇔ $x^{2}$+$x^{2}$-2$x^{2}$+3x-3x-2x=-4-6

⇔ -2x=-10

⇔   x=5(TMĐK)

    Vậy S={5}

d, Ta có: |2x-3|-5=4x

         ⇔  |2x-3|   =4x-5

     ĐK: 4x-5$\geq$0⇔4x≥5⇔x≥$\frac{5}{4}$ 

     Ta có: |2x-3|=4x-5

         ⇔ \(\left[ \begin{array}{l}2x-3=4x-5\\2x-3=-4x+5\end{array} \right.\) 

         ⇔ \(\left[ \begin{array}{l}-2x=-2\\6x=8\end{array} \right.\) 

         ⇔ \(\left[ \begin{array}{l}x=1(loại)\\x=\frac{4}{3}(TMĐK) \end{array} \right.\) 

        Vậy S={$\frac{4}{3}$}