Đáp án: m= 0 hoặc m=4
Giải thích các bước giải:
$\begin{array}{l}
{x^2} = - 4x + {m^2} - 4m\\
\Rightarrow {x^2} + 4x - {m^2} + 4m = 0\\
\Rightarrow \Delta ' > 0\\
\Rightarrow 4 + {m^2} - 4m > 0\\
\Rightarrow {\left( {m - 2} \right)^2} > 0\\
\Rightarrow m \ne 2\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = - 4\\
{x_1}{x_2} = - {m^2} + 4m
\end{array} \right.\\
Do:x_1^2 + 4{x_1} - {m^2} + 4m = 0\\
\Rightarrow x_1^2 + 4{x_1} = {m^2} - 4m\\
\Rightarrow {x_2} = x_1^3 + 4x_1^2\\
\Rightarrow {x_2} = {x_1}\left( {x_1^2 + 4{x_1}} \right)\\
\Rightarrow {x_2} = {x_1}.\left( {{m^2} - 4m} \right)\\
Thay\,vao\,:{x_1} + {x_2} = - 4\\
\Rightarrow {x_1} + {x_1}.\left( {{m^2} - 4m} \right) = - 4\\
\Rightarrow {x_1}\left( {{m^2} - 4m + 1} \right) = - 4\\
\Rightarrow {x_1} = \dfrac{{ - 4}}{{{m^2} - 4m + 1}}\\
\Rightarrow {x_2} = \dfrac{{ - 4\left( {{m^2} - 4m} \right)}}{{{m^2} - 4m + 1}}\\
Thay\,vao:{x_1}.{x_2} = - {m^2} + 4m\\
\Rightarrow \dfrac{{ - 4}}{{{m^2} - 4m + 1}}.\dfrac{{ - 4\left( {{m^2} - 4m} \right)}}{{{m^2} - 4m + 1}} = - {m^2} + 4m\\
\Rightarrow \left( {{m^2} - 4m} \right).\left( {\dfrac{{16}}{{{{\left( {{m^2} - 4m + 1} \right)}^2}}} + 1} \right) = 0\\
\Rightarrow {m^2} - 4m = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 0\left( {tm} \right)\\
m = 4\left( {tm} \right)
\end{array} \right.
\end{array}$
Vậy m= 0 hoặc m=4
