Đáp án + giải thích các bước giải:

` \root{3}{3x+1}+\root{3}{5-x}+\root{3}{2x-9}=\root{3}{4x-3}`

`->3x+1+5-x+2x-9+3(\root{3}{3x+1}+\root{3}{5-x})(\root{3}{5-x}+\root{3}{2x-9})(\root{3}{3x+1}+\root{3}{2x-9})=4x-3 `

`->4x-3+3(\root{3}{3x+1}+\root{3}{5-x})(\root{3}{5-x}+\root{3}{2x-9})(\root{3}{3x+1}+\root{3}{2x-9})=4x-3`

`->3(\root{3}{3x+1}+\root{3}{5-x})(\root{3}{5-x}+\root{3}{2x-9})(\root{3}{3x+1}+\root{3}{2x-9})=0`

`->(\root{3}{3x+1}+\root{3}{5-x})(\root{3}{5-x}+\root{3}{2x-9})(\root{3}{3x+1}+\root{3}{2x-9})=0`

`->`\(\left[ \begin{array}{l}\sqrt[3]{3x+1}+\sqrt[3]{5-x}=0\\\sqrt[3]{5-x}+\sqrt[3]{2x-9}=0\\\sqrt[3]{3x+1}+\sqrt[3]{2x-9}=0\end{array} \right.\) 

`->`\(\left[ \begin{array}{l}3x+1=-(5-x)\\5-x=-(2x-9)\\3x+1=-(2x-9)\end{array} \right.\) 

`->`\(\left[ \begin{array}{l}x=-3\\x=4\\x=\dfrac{8}{5}\end{array} \right.\) 

Vậy `S={-3;4;8/5}`

đặt $\sqrt[3]{3x+1}$ =a

$\sqrt[3]{5-x}$ =b

$\sqrt[3]{2x-9}$ =c

nên a³+b³+c³=3x+1+5-x+2x-9=4x-3

nên phương trình 

⇔a+b+c=$\sqrt[3]{a³+b³+c³}$ 

⇔(a+b+c)³=a³+ b³+c³

⇔a³+ b³+c³=(a+b+c)³=(a+b)³+3(a+b)c(a+b+c)+c³=

=a³+b³+3ab(a+b)+3(a+b)(c²+ca+cb)+c³=(a³+b³+c³)+3(a+b)(ab+ca+cb+c²)

⇔3(a+b)(ab+ca+cb+c²)=0

⇔3(a+b)(b+c)(c+a)=0

⇒a=-b hoặc b=-c hoặc c=-a

nếu a=-b ⇒ $\sqrt[3]{3x+1}$=-$\sqrt[3]{5-x}$ 

                    ⇔($\sqrt[3]{3x+1}$)³=-($\sqrt[3]{5-x}$ )³

                  ⇔3x+1=x+5

⇔2x=4⇔x=2

nếu b=-c thì  -$\sqrt[3]{5-x}$ =$\sqrt[3]{2x-9}$ 

⇔(-$\sqrt[3]{5-x}$)³=($\sqrt[3]{2x-9}$ )³

⇔x-5=2x-9

⇔x=4

nếu c=-a⇔$\sqrt[3]{2x-9}$ =-$\sqrt[3]{3x+1}$

              ⇔(-$\sqrt[3]{2x-9}$)³=($\sqrt[3]{3x+1}$)³

⇔9-2x=3x+1

⇔x=1.6

vậy x={1.6;2;4)