Đáp án:
C2:
b) \(\left[ \begin{array}{l}
m = 1\\
m = - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
a\left( {x + 2} \right)\left( {x + 5} \right)\left( {x + 3} \right)\left( {x + 4} \right) = 24\\
\to \left( {{x^2} + 7x + 10} \right)\left( {{x^2} + 7x + 12} \right) = 24\\
Đặt:{x^2} + 7x + 10 = t\\
\to t\left( {t + 2} \right) = 24\\
\to {t^2} + 2t - 24 = 0\\
\to \left[ \begin{array}{l}
t = 4\\
t = - 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} + 7x + 10 = 4\\
{x^2} + 7x + 10 = - 6\left( {vô nghiệm} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x = - 6
\end{array} \right.\\
b)\left( {x + 1} \right)\left( {x + 5} \right)\left( {x + 2} \right)\left( {x + 4} \right) - 40 = 0\\
\to \left( {{x^2} + 6x + 5} \right)\left( {{x^2} + 6x + 8} \right) - 40 = 0\\
Đặt:{x^2} + 6x + 5 = t\\
Pt \to t\left( {t + 3} \right) - 40 = 0\\
\to {t^2} + 3t - 40 = 0\\
\to \left( {t + 8} \right)\left( {t - 5} \right) = 0\\
\to \left[ \begin{array}{l}
t = - 8\\
t = 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} + 6x + 5 = - 8\left( {vô nghiệm} \right)\\
{x^2} + 6x + 5 = 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 6\\
x = 0
\end{array} \right.\\
C2:\\
a)DK:\Delta > 0\\
\to 4{m^2} - 4m + 1 - 4\left( {{m^2} - 1} \right) > 0\\
\to - 4m + 5 > 0\\
\to \dfrac{5}{4} > m\\
\to \left[ \begin{array}{l}
x = \dfrac{{2m - 1 + \sqrt {5 - 4m} }}{2}\\
x = \dfrac{{2m - 1 - \sqrt {5 - 4m} }}{2}
\end{array} \right.\\
b)DK:\dfrac{5}{4} \ge m\\
{\left( {{x_1} - {x_2}} \right)^2} = {x_1} - 3{x_2}\\
\to {x_1}^2 - 2{x_1}{x_2} + {x_2}^2 = {x_1} - 3{x_2}\\
\to {x_1}^2 + 2{x_1}{x_2} + {x_2}^2 - 4{x_1}{x_2} = {x_1} + {x_2} - 4{x_2}\\
\to {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = {x_1} + {x_2} - 4{x_2}\\
\to \left[ \begin{array}{l}
4{m^2} - 4m + 1 - 4\left( {{m^2} - 1} \right) = 2m - 1 - 4\left( {\dfrac{{2m - 1 + \sqrt {5 - 4m} }}{2}} \right)\\
4{m^2} - 4m + 1 - 4\left( {{m^2} - 1} \right) = 2m - 1 - 4\left( {\dfrac{{2m - 1 - \sqrt {5 - 4m} }}{2}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 4m + 5 = 2m - 1 - 2\left( {2m - 1 + \sqrt {5 - 4m} } \right)\\
- 4m + 5 = 2m - 1 - 2\left( {2m - 1 - \sqrt {5 - 4m} } \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
6m - 6 - 4m + 2 - 2\sqrt {5 - 4m} = 0\\
6m - 6 - 4m + 2 + 2\sqrt {5 - 4m} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
2m - 4 = 2\sqrt {5 - 4m} \\
2\sqrt {5 - 4m} = 4 - 2m
\end{array} \right.\\
\to m - 2 = \sqrt {5 - 4m} \\
\to {m^2} - 4m + 4 = 5 - 4m\\
\to {m^2} = 1\\
\to \left[ \begin{array}{l}
m = 1\\
m = - 1
\end{array} \right.\left( {TM} \right)
\end{array}\)
