Đáp án:
b) \(\left[ \begin{array}{l}
m = 1\\
m = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
C3:\\
1)DK:x \ne - 2;y \ne \dfrac{1}{2}\\
\left\{ \begin{array}{l}
\dfrac{1}{{x + 2}} + \dfrac{3}{{2y - 1}} = 4\\
\dfrac{{12}}{{x + 2}} - \dfrac{3}{{2y - 1}} = 9
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{13}}{{x + 2}} = 13\\
\dfrac{1}{{x + 2}} + \dfrac{3}{{2y - 1}} = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x + 2 = 1\\
2y - 1 = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = - 1\\
y = 1
\end{array} \right.\\
2)a)Thay:m = 2\\
Pt \to {x^2} - 4x + 3 = 0\\
\to \left( {x - 3} \right)\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\\
b)Xet:\Delta ' \ge 0\\
\to {m^2} - 2m + 1 \ge 0\\
\to {\left( {m - 1} \right)^2} \ge 0\forall m\\
Có:{x_1}^2 + {x_2}^2 = 10\\
\to \left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2}} \right) - 2{x_1}{x_2} = 10\\
\to {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 10\\
\to 4{m^2} - 2\left( {2m - 1} \right) = 10\\
\to 4{m^2} - 4m + 2 = 10\\
\to \left[ \begin{array}{l}
m = 1\\
m = - 2
\end{array} \right.
\end{array}\)
