Đáp án:
c) m=7
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:\left\{ \begin{array}{l}
{m^2} - 4m + 4 - 4\left( {m + 5} \right) \ge 0\\
{x_1}^2 + {x_2}^2 = 10
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{m^2} - 4m + 4 - 4m - 20 \ge 0\\
{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2} = 10
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{m^2} - 8m - 16 \ge 0\\
{\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 10
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{\left( {m - 4} \right)^2} \ge 0\left( {ld} \right)\forall m\\
{\left( {2 - m} \right)^2} - 2\left( {m + 5} \right) = 10
\end{array} \right.\\
\to 4 - 4m + {m^2} - 2m - 10 = 10\\
\to {m^2} - 6m - 16 = 0\\
\to \left( {m - 8} \right)\left( {m + 2} \right) = 0\\
\to \left[ \begin{array}{l}
m = 8\\
m = - 2
\end{array} \right.\\
b)DK:\left\{ \begin{array}{l}
{m^2} - 4\left( {m + 1} \right) \ge 0\\
{x_1}{x_2} + 2\left( {{x_1} + {x_2}} \right) = 19
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{m^2} - 4m - 4 \ge 0\\
m + 1 + 2m = 19
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m \ge 2 + 2\sqrt 2 \\
m \le 2 - 2\sqrt 2
\end{array} \right.\\
3m = 18
\end{array} \right.\\
\to m = 6\left( {TM} \right)\\
c)DK:\left\{ \begin{array}{l}
m \ne - 1\\
{m^2} + 4m + 4 - \left( {m + 1} \right)\left( {m - 3} \right) \ge 0\\
\left( {4{x_1} + 1} \right)\left( {4{x_2} + 1} \right) = 18
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne - 1\\
{m^2} + 4m + 4 - {m^2} + 2m + 3 \ge 0\\
16{x_1}{x_2} + 4\left( {{x_1} + {x_2}} \right) + 1 = 18
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne - 1\\
m \ge - \dfrac{7}{6}\\
16.\dfrac{{m - 3}}{{m + 1}} + 4.\dfrac{{2m + 4}}{{m + 1}} = 17
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne - 1\\
m \ge - \dfrac{7}{6}\\
16m - 16.3 + 8m + 16 = 17m + 17
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne - 1\\
m \ge - \dfrac{7}{6}\\
m = 7\left( {TM} \right)
\end{array} \right.
\end{array}\)
