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={$\frac{4}{3}$}
