Đáp án: $m = \frac{1}{4}$
Giải thích các bước giải:
$\begin{array}{l}
b){x^2} - \left( {m + 1} \right)x + m = 0\\
\Delta > 0\\
\Leftrightarrow {\left( {m + 1} \right)^2} - 4m > 0\\
\Leftrightarrow {m^2} - 2m + 1 > 0\\
\Leftrightarrow {\left( {m - 1} \right)^2} > 0\\
\Leftrightarrow m\# 1\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m + 1\\
{x_1}{x_2} = m
\end{array} \right.\\
\left\{ \begin{array}{l}
x_1^2 - \left( {m + 1} \right){x_1} + m = 0\\
x_2^2 - \left( {m + 1} \right){x_2} + m = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x_1^2 - m{x_1} = {x_1} - m\\
x_2^2 - m{x_2} = {x_2} - m
\end{array} \right.\\
Khi:\left( {x_1^2 - m{x_1} + {x_2} + 2m} \right)\\
.\left( {x_2^2 - m{x_2} + {x_1} + 2m} \right) = 9{x_1}{x_2}\\
\Leftrightarrow \left( {{x_1} - m + {x_2} + 2m} \right)\left( {{x_2} - m + {x_1} + 2m} \right) = 9{x_1}{x_2}\\
\Leftrightarrow \left( {m + 1 + m} \right)\left( {m + 1 + m} \right) = 9m\\
\Leftrightarrow {\left( {2m + 1} \right)^2} = 9m\\
\Leftrightarrow 4{m^2} + 4m + 1 = 9m\\
\Leftrightarrow 4{m^2} - 5m + 1 = 0\\
\Leftrightarrow \left( {4m - 1} \right)\left( {m - 1} \right) = 0\\
\Leftrightarrow m = \frac{1}{4}\left( {do:m\# 1} \right)\\
Vậy\,m = \frac{1}{4}
\end{array}$
