Đáp án:
\(4{x_1}{x_2} + \left( {{x_1} + {x_2}} \right) - 2 = 0.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left( {m - 4} \right){x^2} - 2\left( {m - 2} \right)x + m - 1 = 0\,\,\,\,\left( * \right)\\
\left( * \right)\,\,\,co\,\,\,2\,\,\,nghiem\,\,phan\,\,biet\,\,\,{x_1},\,\,{x_2}\\
\Leftrightarrow \left\{ \begin{array}{l}
a \ne 0\\
\Delta ' > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m - 4 \ne 0\\
{\left( {m - 2} \right)^2} - \left( {m - 4} \right)\left( {m - 1} \right) > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne 0\\
{m^2} - 4m + 4 - {m^2} + 5m - 4 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m \ne 0\\
m > 0
\end{array} \right. \Leftrightarrow m > 0.\\
Ap\,\,dung\,\,dinh\,\,ly\,\,Vi - et\,\,ta\,\,co:\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{2\left( {m - 2} \right)}}{{m - 4}}\\
{x_1}{x_2} = \frac{{m - 1}}{{m - 4}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left( {m - 4} \right)\left( {{x_1} + {x_2}} \right) = 2m - 4\\
m{x_1}{x_2} - 4{x_1}{x_2} = m - 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m\left( {{x_1} + {x_2}} \right) - 4\left( {{x_1} + {x_2}} \right) = 2m - 4\\
m\left( {{x_1}{x_2} - 1} \right) = 4{x_1}{x_2} - 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m\left( {{x_1} + {x_2} - 2} \right) = 4\left( {{x_1} + {x_2}} \right) - 4\\
m = \frac{{4{x_1}{x_2} - 1}}{{{x_1}{x_2}}}
\end{array} \right.\\
\Rightarrow \frac{{4\left( {{x_1} + {x_2} - 1} \right)}}{{{x_1} + {x_2} - 2}} = \frac{{4{x_1}{x_2} - 1}}{{{x_1}{x_2}}}\\
\Leftrightarrow 4\left( {{x_1} + {x_2} - 1} \right){x_1}{x_2} = \left( {4{x_1}{x_2} - 1} \right)\left( {{x_1} + {x_2} - 2} \right)\\
\Leftrightarrow 4\left( {{x_1} + {x_2}} \right){x_1}{x_2} - 4{x_1}{x_2} = 4\left( {{x_1} + {x_2}} \right){x_1}{x_2} - 8{x_1}{x_2} - \left( {{x_1} + {x_2}} \right) + 2\\
\Leftrightarrow 4{x_1}{x_2} + \left( {{x_1} + {x_2}} \right) - 2 = 0.
\end{array}\)
