Đáp án:
C3:
b) m=11
Giải thích các bước giải:
\(\begin{array}{l}
C2:\\
a)Thay:x = 1\\
\left( 1 \right) \to 1 + 4 + m = 0\\
\to m = - 5\\
b)DK:\Delta ' = 0\\
\to 4 - m = 0\\
\to m = 4\\
c)Thay:m = - 12\\
\left( 1 \right) \to {x^2} + 4x - 12 = 0\\
\to \left( {x + 6} \right)\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 6\\
x = 2
\end{array} \right.\\
C3:\\
a)Thay:m = 1\\
\left( 1 \right) \to {x^2} + x - 1 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 1 + \sqrt 5 }}{2}\\
x = \dfrac{{ - 1 - \sqrt 5 }}{2}
\end{array} \right.\\
b)DK:\Delta \ge 0\\
\to 4{m^2} - 4m + 1 - 4m \ge 0\\
\to 4{m^2} - 16m + 1 \ge 0\\
\to \left[ \begin{array}{l}
m \ge \dfrac{{4 + \sqrt {15} }}{2}\\
m \le \dfrac{{4 - \sqrt {15} }}{2}
\end{array} \right.\\
Có:2\left( {{x_1} + {x_2}} \right) - 3{x_1}{x_2} + 9 = 0\\
\to 2\left( {1 - 2m} \right) - 3\left( { - m} \right) + 9 = 0\\
\to - m + 11 = 0\\
\to m = 11\left( {TM} \right)
\end{array}\)