`a,(x^2-9)-(x+2)(x+3)=0`

`⇔(x-3)(x+3)-(x+2)(x+3)=0`

`⇔(x+3)(x-3-x-2)=0`

`⇔(x+3).(-5)=0`

`⇔x+3=0`

`⇔x=-3`

 Vậy: `S={-3}`

`b,(2x-7)^2=2x-7`

`⇔(2x-7)^2-(2x-7)=0`

`⇔(2x-7)(2x-7-1)=0`

`⇔(2x-7)(2x-8)=0`

`⇔`\(\left[ \begin{array}{l}2x-7=0\\2x-8=0\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}2x=7\\2x=8\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=4\end{array} \right.\) 

Vậy: `S={\frac{7}{2};4}`

`c,3x^3-7x=0`

`⇔x(3x^2-7)=0`

`⇔`\(\left[ \begin{array}{l}x=0\\3x^2-7=0\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=0\\3x^2=7\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=0\\x^2=\dfrac{7}{3}\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=0\\x=\sqrt{\dfrac{7}{3}}\\x=-\sqrt{\dfrac{7}{3}}\end{array} \right.\) 

Vậy: `S={0;\sqrt{\frac{7}{3}};-\sqrt{\frac{7}{3}}}`

Đáp án:

 `a, (x^2 - 9) - (x + 2)(x + 3) = 0`

`⇔ (x^2 - 3^2) - (x + 2)(x + 3) = 0`

`⇔ (x - 3)(x + 3) - (x + 2)(x + 3) = 0`

`⇔ (x + 3)(x - 3 - x - 2) = 0`

`⇔ (x + 3). (-5) = 0`

`⇔ x + 3 = 0`

`⇒ x =0 - 3`

`⇒ x = -3`

Vậy `x = -3`

`b, (2x - 7)^2 = (2x - 7)`

`⇔ (2x - 7)^2 - (2x - 7) = 0`

`⇔ (2x - 7)(2x - 7 - 1) = 0`

`⇔ (2x - 7)(2x - 8) = 0`

`⇔ (2x - 7)(x - 4).2 = 0`

`⇒` $\left[\begin{matrix} 2x - 7 = 0\\ x - 4 = 0\end{matrix}\right.$

`⇒` $\left[\begin{matrix} 2x = 7\\ x = 4\end{matrix}\right.$

`⇒` $\left[\begin{matrix} x = \frac{7}{2}\\ x = 4\end{matrix}\right.$

Vậy `x = 7/2` hoặc `x = 4`

`c, 3x^3 - 7x = 0`

`⇔ x(3x^2 - 7) = 0`

`⇒` $\left[\begin{matrix} x = 0\\ 3x^2 - 7 = 0\end{matrix}\right.$

`⇒` $\left[\begin{matrix} x = 0\\ 3x^2 = 7\end{matrix}\right.$

`⇒` $\left[\begin{matrix} x = 0\\ x^2 = \dfrac{7}{3}\end{matrix}\right.$

`⇒` $\left[\begin{matrix} x = 0\\ x^2 = (\sqrt{\dfrac{7}{3}})^2\\x^2 = (-\sqrt{\dfrac{7}{3}})^2\end{matrix}\right.$

`⇒` $\left[\begin{matrix} x = 0\\ x = \sqrt{\dfrac{7}{3}}\\x^2 = -\sqrt{\dfrac{7}{3}}\end{matrix}\right.$

Vậy `x = 0, x =`$\sqrt{\dfrac{7}{3}}$ hoặc `x =` $-\sqrt{\dfrac{7}{3}}$