Đáp án:
\(\left[ \begin{array}{l}
m = \dfrac{1}{8}\\
m = - \dfrac{1}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:m = 2\\
Pt \to {x^2} - 12x + 11 = 0\\
\to \left( {x - 11} \right)\left( {x - 1} \right) = 0\\
\to \left[ {\begin{array}{*{20}{l}}
{x = 11}\\
{x = 1}
\end{array}} \right.\\
b)\Delta ' = 4\left( {{m^2} + 2m + 1} \right) - 4{m^2} + 2m + 1\\
= 4{m^2} + 8m + 4 - 4{m^2} + 2m + 1\\
= 10m + 5
\end{array}\)
Để phương trình có nghiệm
\(\begin{array}{l}
\to \Delta ' \ge 0\\
\to 10m + 5 \ge 0\\
\to m \ge - \dfrac{1}{2}\\
\to \left[ {\begin{array}{*{20}{l}}
{x = 2\left( {m + 1} \right) + \sqrt {10m + 5} }\\
{x = 2\left( {m + 1} \right) - \sqrt {10m + 5} {\rm{\;}}}
\end{array}} \right.\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 4m + 4\\
{x_1}{x_2} = 4{m^2} - 2m - 1
\end{array} \right.
\end{array}\)
Theo định lý Vi-et:
\(\begin{array}{l}
Có:{x_1} + 2{x_2} + {x_1}{x_2} = 4{m^2} + 3\\
\to {x_1} + {x_2} + {x_2} + {x_1}{x_2} = 4{m^2} + 3\\
\to \left[ {\begin{array}{*{20}{l}}
{4m + 4 + 2\left( {m + 1} \right) + \sqrt {10m + 5} {\rm{ \;}} + 4{m^2} - 2m - 1 = 4{m^2} + 3}\\
{4m + 4 + 2\left( {m + 1} \right) - \sqrt {10m + 5} {\rm{ \;}} + 4{m^2} - 2m - 1 = 4{m^2} + 3}
\end{array}} \right.\\
\to \left[ {\begin{array}{*{20}{l}}
{6m + 6 + \sqrt {10m + 5} {\rm{ \;}} = 2m + 4}\\
{6m + 6 - \sqrt {10m + 5} {\rm{ \;}} = 2m + 4}
\end{array}} \right.\\
\to \sqrt {10m + 5} {\rm{ \;}} = 4m + 2\\
\to 10m + 5 = 16{m^2} + 16m + 4\\
\to 16{m^2} + 6m - 1 = 0\\
\to \left[ {\begin{array}{*{20}{l}}
{m = \dfrac{1}{8}}\\
{m = {\rm{ \;}} - \dfrac{1}{2}}
\end{array}} \right.
\end{array}\)
