Đáp án`+`Giải thích các bước giải:

 `1, (2x + 7)(x - 1) = 0`

`<=>` $\left[\begin{matrix} 2x+ 7 = 0\\ x-1=0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=\dfrac{-7}{2}\\ x=1\end{matrix}\right.$

Vậy `S= {-7/2;1}`

`2, (x - 1)(3x-  6) = 0`

`<=>` $\left[\begin{matrix} x-1=0\\ 3x- 6 =0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=1\\ 3x = 6\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=1\\ x=2\end{matrix}\right.$

Vậy `S= {1;2}`

`3, (2x + 5)(1 - 3x) = 0`

`<=>` $\left[\begin{matrix} 2x + 5= 0\\ 1 - 3x= 0 \end{matrix}\right.$

`<=>` $\left[\begin{matrix} x= \dfrac{-5}{2}\\ x=\dfrac{1}{3}\end{matrix}\right.$

Vậy `S= {-5/2;1/3}`

`4, x^2 - 2x=  0`

`<=> x(x - 2) = 0`

`<=>` $\left[\begin{matrix} x=0\\ x-2 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=0\\ x=2\end{matrix}\right.$

Vậy `S= {0;2}`

`5, 3x^2- 6x=0`

`<=> 3x(x - 2) = 0`

`<=>` $\left[\begin{matrix} 3x= 0\\ x-2= 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=0\\ x=2\end{matrix}\right.$

Vậy `S= {0;2}`

`6, (5 -x)^2 ( 3x - 1) =0`

`<=>` $\left[\begin{matrix} (5 -x)^2= 0\\ 3x-1=0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} 5 -x = 0\\ 3x=1\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x= 5\\ x=\dfrac{1}{3}\end{matrix}\right.$

Vậy `S= {5;1/3}`

`7, (x + 1)^2 (x + 2) =0`

`<=>` $\left[\begin{matrix} (x + 1)^2 = 0\\ x+2 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x+ 1= 0\\ x=-2\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=-1\\ x=-2\end{matrix}\right.$

Vậy `S= {-1;-2}`

`8, x(x + 1) (x + 3/4) = 0`

`<=>` $\left[\begin{matrix} x= 0\\ x+1 = 0\\x + \dfrac{3}{4} = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=0\\ x=-1\\x = \dfrac{-3}{4}\end{matrix}\right.$

Vậy `S= {0;-1;-3/4}`

`***`sharksosad