Đáp án:
B5:
a) \(\dfrac{1}{{12}} \ge m\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)Thay:m = 0\\
Pt \to {x^2} - 5x + 4 = 0\\
\to \left( {x - 1} \right)\left( {x - 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = 4
\end{array} \right.\\
Thay:m = 2\\
Pt \to {x^2} - 5x + 6 = 0\\
\to \left( {x - 3} \right)\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = 2
\end{array} \right.\\
b)Xét:\Delta > 0\\
\to 25 - 4\left( {m + 4} \right) > 0\\
\to 25 - 4m - 16 > 0\\
\to 9 - 4m > 0\\
\to \dfrac{9}{4} > m\\
Có:{x_1}^2 + {x_2}^2 = 2\\
\to {x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2} = 2\\
\to {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 2\\
\to 25 - 2\left( {m + 4} \right) = 2\\
\to 25 - 2m - 8 = 2\\
\to m = \dfrac{{15}}{2}\left( {KTM} \right)\\
\to m \in \emptyset \\
B5:\\
a)DK:\Delta \ge 0\\
\to 4{m^2} - 4m + 1 - 4m\left( {m + 2} \right) \ge 0\\
\to 4{m^2} - 4m + 1 - 4{m^2} - 8m \ge 0\\
\to - 12m + 1 \ge 0\\
\to \dfrac{1}{{12}} \ge m\\
\to \left[ \begin{array}{l}
x = \dfrac{{2m - 1 + \sqrt {1 - 12m} }}{{2m}}\\
x = \dfrac{{2m - 1 - \sqrt {1 - 12m} }}{{2m}}
\end{array} \right.
\end{array}\)
( câu b bị mờ với thiếu đề b nha )
