Tìm m để PT có 2 no PB :
x1^3 - x2^3 +9.x1.x2 =81
\[\begin{align}
& {{x}^{2}}-3x+m-2=0 \\
& \vartriangle =9-4(m-2)=1-4m\ge 0\Rightarrow m\le \frac{1}{4} \\
& Viet:{{x}_{1}}+{{x}_{2}}=3;{{x}_{1}}.{{x}_{2}}=m-2 \\
& {{x}_{1}}^{3}-{{x}_{2}}^{3}+9.{{x}_{1}}.{{x}_{2}}=81 \\
& ({{x}_{1}}-{{x}_{2}})(x_{1}^{2}+{{x}_{1}}.{{x}_{2}}+x_{2}^{2})+9(m-2)=81 \\
& ({{x}_{1}}-{{x}_{2}})\left[ {{({{x}_{1}}+{{x}_{2}})}^{2}}-{{x}_{1}}.{{x}_{2}} \right]+9m-18=81 \\
& ({{x}_{1}}-{{x}_{2}})\left( 11-m \right)=99-9m \\
& {{x}_{1}}-{{x}_{2}}=9 \\
& {{x}_{1}}+{{x}_{2}}=3 \\
& \Rightarrow {{x}_{1}}=...;{{x}_{2}}=...\Rightarrow m=... \\
\end{align}\]