Bạn tham khảo nhé.
`(x+1)(x+2)(x+3)(x+4)=24`
`<=>(x+1)(x+4)(x+2)(x+3)=24`
`<=>(x^2+4x+x+4)(x^2+3x+2x+6)=24`
`<=>(x^2+5x+4)(x^2+5x+6)-24=0` `(1)`
`\text{Đặt}` `x^2+5x+5=a`
`\text{Phương trình}` `(1)` `\text{viết lại:}`
`<=>(a+1)(a-1)-24=0`
`<=>a^2-1^2-24=0`
`<=>a^2-1-24=0`
`<=>a^2-25=0`
`<=>a^2-5^2=0`
`<=>(a-5)(a+5)=0`
`=>(x^2+5x+5-5)(x^2+5x+5+5)=0`
`<=>(x^2+5x)(x^2+5x+10)=0`
`<=>x(x+5)(x^2+2.x. 5/2+25/4+15/4)=0`
`<=>x(x+5)[(x+5/2)^2+15/4]=0`
`<=>`$\left[\begin{matrix} x=0 \\ x+5=0 \\ (x+\dfrac{5}{2})^2+\dfrac{15}{4}=0\end{matrix}\right.$
`<=>`$\left[\begin{matrix} x=0 \\ x=-5 \\ (x+\dfrac{5}{2})^2=-\dfrac{15}{4}(\text{Vô lý})\end{matrix}\right.$
Vậy `x\in{0;-5}`
