\(\begin{array}{l}
{x^2} - 2mx + 4m - 3 = 0\\
Pt\,\,co\,\,2\,\,\,nghiem\,\,pb \Leftrightarrow \Delta ' > 0\\
\Leftrightarrow {m^2} - 4m + 3 > 0\\
\Leftrightarrow \left( {m - 3} \right)\left( {m - 1} \right) > 0 \Leftrightarrow \left[ \begin{array}{l}
m > 3\\
m < 1
\end{array} \right..\\
Ap\,\,dung\,\,\,dinh\,\,ly\,\,Vi - et\,\,ta\,\,co:\,\,\,\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m\\
{x_1}{x_2} = 4m - 3
\end{array} \right..\\
a)\,\,\,{x_1} > 1 > {x_2}\\
\Rightarrow \left( {{x_1} - 1} \right)\left( {{x_2} - 1} \right) < 0\\
\Leftrightarrow {x_1}{x_2} - \left( {{x_1} + {x_2}} \right) + 1 < 0\\
\Leftrightarrow 4m - 3 - 2m + 1 < 0\\
\Leftrightarrow 2m < 2\\
\Leftrightarrow m < 1.\\
Ket\,\,hop\,\,\,dk\,\,\,ta\,\,duoc\,\,m < 1.\\
b)\,\,\,\left\{ \begin{array}{l}
{x_1} > 2\\
{x_2} > 2
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} > 4\\
\left( {{x_1} - 2} \right)\left( {{x_2} - 2} \right) > 0
\end{array} \right.
\end{array}\)
Đến đây em áp dụng định lý Vi-ét làm tiếp nhé em.
