`(x+1)(x+3)(x+5)(x+7)+15=0`

`(x+1)(x+7)(x+3)(x+5)+15=0`

`(x^2+x+7x+7)(x^2+3x+5x+15)+15=0`

`(x^2+8x+7)(x^2+8x+15)+15=0`

`(x^2+8x+7)(x^2+8x+7+8)+15=0`

Đặt `x^2+8x+7=t`, ta có:

`t(t+8)+15=0`

`t^2+8t+15=0`

`t^2+3t+5t+15=0`

`t(t+3)+5(t+3)=0`

`(t+5)(t+3)=0`

`t=-5` hoặc `t=-3`

Thay `t=-5`, ta có:

`x^2+8x+7=-5`

`x^2+8x+12=0`

`x^2+6x+2x+12=0`

`x(x+6)+2(x+6)=0`

`(x+2)(x+6)=0`

`x=-2` hoặc `x=-6`

Thay `t=-3`, ta có:

`x^2+8x+7=-3`

`x^2+8x+10=0`

`x^2+8x+16-6=0`

`(x+4)^2=6`

`x+4=\sqrt{6}` hoặc `x+4=-\sqrt{6}`

`x=\sqrt{6}-4` hoặc `x=-\sqrt{6}-4`

Vậy `x in {-2;-6;\sqrt{6}-4;-\sqrt{6}-4}`