Đáp án:
b. \(\left[ \begin{array}{l}
m = 1\\
m = - 4
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.Thay:m = \sqrt 2 \\
Pt \to {x^2} - \left( {2\sqrt 2 + 3} \right)x - 2\sqrt 2 - 4 = 0\\
Xét: Δ= {\left( {2\sqrt 2 + 3} \right)^2} + 4\left( {2\sqrt 2 + 4} \right) = 33 + 20\sqrt 2 > 0\\
\to \left[ \begin{array}{l}
x = \frac{{2\sqrt 2 + 3 + \sqrt {33 + 20\sqrt 2 } }}{2}\\
x = \frac{{2\sqrt 2 + 3 - \sqrt {33 + 20\sqrt 2 } }}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \frac{{2\sqrt 2 + 3 + \sqrt {33 + 20\sqrt 2 } }}{2}\\
x = - 1
\end{array} \right.
\end{array}\)
b. Phương trình có hai nghiệm phân biệt
⇔ Δ>0
\(\begin{array}{l}
\to 4{m^2} + 12m + 9 + 4\left( {2m + 4} \right) > 0\\
\to 4{m^2} + 12m + 9 + 8m + 16 > 0\\
\to 4{m^2} + 20m + 25 > 0\\
\to {\left( {2m + 5} \right)^2} > 0\\
\Leftrightarrow m \ne \frac{5}{2}\\
Có:\left| {{x_1}} \right| + \left| {{x_2}} \right| = 5\\
\to {x_1}^2 + 2{x_1}{x_2} + {x_2}^2 = 25\\
\to \left( {{x_1}^2 + {x_2}^2} \right) + 2{x_1}{x_2} = 25\\
\to \left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2}} \right) - 2{x_1}{x_2} + 2{x_1}{x_2} = 25\\
\to {\left( {{x_1} + {x_2}} \right)^2} = 25\\
\to {\left( {2m + 3} \right)^2} = 25\\
\to \left[ \begin{array}{l}
2m + 3 = 5\\
2m + 3 = - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = 1\\
m = - 4
\end{array} \right.
\end{array}\)
