giúp tui huhu

$\displaystyle x^{2} -( 2m+1) x+m^{2} +m-6=0$
Ta có: 
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Delta =( 2m+1)^{2} -4.\left( m^{2} +m-6\right)\\
\Delta =4m^{2} +4m+1-4m^{2} -4m+24=25 >0
\end{array}$.
$\displaystyle \Rightarrow $Phương trình luôn có 2 nghiệm phân biệt $\displaystyle x_{1} ;\ x_{2}$
Áp dụng định lý Viette ta có: 
$\displaystyle \begin{cases}
x_{1} +x_{2} =2m+1 & \\
x_{1} .x_{2} =m^{2} +m-6 & 
\end{cases} \Rightarrow \begin{cases}
x_{1} =m-2 & \\
x_{2} =m+3 & 
\end{cases}$
Ta có: $\displaystyle |x_{1}^{3} -x_{2}^{3} |=50\Rightarrow \left[ \begin{array}{l l}
x_{1}^{3} -x_{2}^{3} =50 & \\
x_{1}^{3} -x_{2}^{3} =-50 & 
\end{array} \right.$
+ TH1: $\displaystyle x_{1}^{3} -x_{2}^{3} =50$.
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Rightarrow ( x_{1} -x_{2})\left( x_{1}^{2} +x_{1} x_{2} +x_{2}^{2}\right) =50\\
\Leftrightarrow ( m-2-m-3)\left[( x_{1} +x_{2})^{2} -x_{1} x_{2}\right] =50\\
\Leftrightarrow -5.\left[( 2m+1)^{2} -\left( m^{2} +m-6\right)\right] =50\\
\Leftrightarrow 4m^{2} +4m+1-m^{2} -m+6=-10\\
\Leftrightarrow 3m^{2} +3m+17=0
\end{array}$
Ta có: $\displaystyle \Delta =3^{2} -4.3.17< 0\Rightarrow $Phương trình vô nghiệm 
+ TH2: $\displaystyle x_{1}^{3} -x_{2}^{3} =-50$.
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Rightarrow ( x_{1} -x_{2})\left( x_{1}^{2} +x_{1} .x_{2} +x_{2}^{2}\right) =-50\\
\Leftrightarrow ( m-2-m-3)\left[( x_{1} +x_{2})^{2} -x_{1} x_{2}\right] =-50\\
\Leftrightarrow -5.\left[( 2m+1)^{2} -\left( m^{2} +m-6\right)\right] =-50\\
\Leftrightarrow 4m^{2} +4m+1-m^{2} -m+6=10\\
\Leftrightarrow 3m^{2} +3m-3=0\\
\Leftrightarrow m^{2} +m-1=0\\
\Leftrightarrow \left( m+\frac{1}{2}\right)^{2} =\frac{5}{4}\\
\Leftrightarrow \left[ \begin{array}{l l}
m=\frac{-1-\sqrt{5}}{2} & \\
m=\frac{-1+\sqrt{5}}{2} & 
\end{array} \right.
\end{array}$