Đáp án:
d) \(\left[ \begin{array}{l}
m = - 6 + 3\sqrt 3 \\
m = - 6 - 3\sqrt 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:m + 3 \ne 0\\
\to m \ne - 3\\
b)Thay:m = 2\\
\left( 1 \right) \to 5{x^2} + 4x - 1 = 0\\
\to \left( {5x - 1} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{5}\\
x = - 1
\end{array} \right.\\
c)DK:\left( {m + 3} \right)\left( {m - 3} \right) < 0\\
\to {m^2} - 9 < 0\\
\to {m^2} < 9\\
\to - 3 < m < 3\\
d)Xét:\Delta ' \ge 0\\
\to {m^2} - \left( {m + 3} \right)\left( {m - 3} \right) \ge 0\\
\to 9 \ge 0\left( {ld} \right)\\
Có:{x_1}^2 + {x_2}^2 = 4\\
\to \left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2}} \right) - 2{x_1}{x_2} = 4\\
\to {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 4\\
\to {\left( { - \dfrac{{2m}}{{m + 3}}} \right)^2} - 2.\dfrac{{m - 3}}{{m + 3}} = 4\\
\to \dfrac{{4{m^2} - 2\left( {m - 3} \right)\left( {m + 3} \right) - 4{{\left( {m + 3} \right)}^2}}}{{{{\left( {m + 3} \right)}^2}}} = 0\\
\to 4{m^2} - 2{m^2} + 18 - 4\left( {{m^2} + 6m + 9} \right) = 0\\
\to 4{m^2} - 2{m^2} + 18 - 4{m^2} - 24m - 36 = 0\\
\to - 2{m^2} - 24m - 18 = 0\\
\to \left[ \begin{array}{l}
m = - 6 + 3\sqrt 3 \\
m = - 6 - 3\sqrt 3
\end{array} \right.
\end{array}\)
