Đáp án:
\(a)\left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.\)
b) m=-5
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:m = 3\\
Pt \to {x^2} - 2x = 0\\
\to x\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.
\end{array}\)
b) Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to \Delta ' > 0\\
\to 1 - m + 3 > 0\\
\to 4 > m\\
\to \left[ \begin{array}{l}
x = 1 + \sqrt {4 - m} \\
x = 1 - \sqrt {4 - m}
\end{array} \right.\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = m - 3
\end{array} \right.\\
{x_1}^2 - 2{x_2} + {x_1}{x_2} = - 12\\
\to {x_1}^2 - \left( {{x_1} + {x_2}} \right){x_2} + {x_1}{x_2} = - 12\\
\to {x_1}^2 - {x_1}{x_2} - {x_2}^2 + {x_1}{x_2} = - 12\\
\to \left( {{x_1} - {x_2}} \right)\left( {{x_1} + {x_2}} \right) = - 12\\
\to 2\left( {{x_1} - {x_2}} \right) = - 12\\
\to {x_1} - {x_2} = - 6\\
\to \left[ \begin{array}{l}
1 + \sqrt {4 - m} - \left( {1 - \sqrt {4 - m} } \right) = - 6\\
1 - \sqrt {4 - m} - \left( {1 + \sqrt {4 - m} } \right) = - 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
2\sqrt {4 - m} = - 6\left( l \right)\\
2\sqrt {4 - m} = 6
\end{array} \right.\\
\to \sqrt {4 - m} = 3\\
\to 4 - m = 9\\
\to m = - 5
\end{array}\)
