Đáp án:
a) \(\left[ \begin{array}{l}
x = 1\\
x = 5
\end{array} \right.\)
Giải thích các bước giải:
Bài 2:
\(\begin{array}{l}
a)Thay:m = 1\\
Pt \to {x^2} - 6x + 5 = 0\\
\to \left( {x - 1} \right)\left( {x - 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = 5
\end{array} \right.\\
b)Thay:x = - 2\\
Pt \to 4 + 2\left( {m + 5} \right) - m + 6 = 0\\
\to 2m + 10 - m + 10 = 0\\
\to m + 20 = 0\\
\to m = - 20\\
c)DK:\Delta \ge 0\\
\to {m^2} + 10m + 25 - 4\left( { - m + 6} \right) \ge 0\\
\to {m^2} + 10m + 25 + 4m - 24 \ge 0\\
\to {m^2} + 14m + 1 \ge 0\\
\to \left[ \begin{array}{l}
m \ge - 7 + 4\sqrt 3 \\
m \le - 7 - 4\sqrt 3
\end{array} \right.\\
{x_1}^2{x_2} + {x_1}{x_2}^2 = 24\\
\to {x_1}{x_2}\left( {{x_1} + {x_2}} \right) = 24\\
\to \left( { - m + 6} \right)\left( {m + 5} \right) = 24\\
\to - {m^2} + m + 30 = 24\\
\to - {m^2} + m + 6 = 0\\
\to \left( {3 - m} \right)\left( {m + 2} \right) = 0\\
\to \left[ \begin{array}{l}
m = 3\\
m = - 2\left( l \right)
\end{array} \right.
\end{array}\)
Bài 3:
\(\begin{array}{l}
a)Thay:m = 1\\
Pt \to {x^2} - 4x + 3 = 0\\
\to \left( {x - 1} \right)\left( {x - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.\\
b)Xét:\Delta ' \ge 0\\
\to {m^2} + 2m + 1 - {m^2} - 2m \ge 0\\
\to 1 \ge 0\left( {ld} \right)\\
{x_1}^2 + {x_2}^2 = {x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}\\
= {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2}\\
= {\left( {2m + 2} \right)^2} - 2\left( {{m^2} + 2m} \right)\\
= 4{m^2} + 8m + 4 - 2{m^2} - 4m\\
= 2{m^2} + 4m + 4\\
= 2\left( {{m^2} + 2m + 2} \right)\\
= 2{\left( {m + 1} \right)^2} + 2\\
Do:2{\left( {m + 1} \right)^2} \ge 0\left( {ld} \right)\forall m\\
\to 2{\left( {m + 1} \right)^2} + 2 \ge 2\\
\to Min = 2\\
\Leftrightarrow m + 1 = 0\\
\to m = - 1
\end{array}\)
