`e) (x + 1).(2x - 7) = 0`

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

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

Vậy phương trình có hai nghiệm `(-1; 7/2)`

`g) \frac{x+7}{3} - \frac{2x+5}{4} = x/6`

`<=> (x + 7).4 - 3.(2x + 5) = 2x`

`<=> 4x + 28 - 6x - 15 = 2x`

`<=> 4x + 28 - 6x - 15 - 2x = 0`

`<=> -4x + 13 = 0`

`<=> -4x = -13`

`<=> x = (-13):(-4)`

`<=> x = 13/4`

Vậy `x = 13/4`

`h) \frac{x-2}{2} - \frac{x-3}{3} = \frac{x+10}{12}`

`<=> 6.(x - 2) - 4.(x - 3) = x + 10`

`<=> 6x - 12 - 4x + 12 - x - 10 = 0`

`<=> x - 10 = 0`

`<=> x = 10`

Vậy `x = 10`

`i) (x + 2)^2 - (x - 3)^2 = 15`

`<=> x^2 + 4x + 4 - (x^2 - 6x + 9) = 15`

`<=> x^2 + 4x + 4 - x^2 + 6x - 9 = 15`

`<=> 10x - 5 = 15`

`<=> 10x = 20`

`<=> x = 20:10`

`<=> x = 2`

Vậy `x = 2`