Đáp án:
a) \(m > - \dfrac{7}{8}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Ycbt \Leftrightarrow 1 - 8m + 16{m^2} - 8\left( {2{m^2} + 1} \right) < 0\\
\to 1 - 8m + 16{m^2} - 16{m^2} - 8 < 0\\
\to - 7 < 8m\\
\to m > - \dfrac{7}{8}\\
b)Ycbt \Leftrightarrow 1 - 8m + 16{m^2} - 8\left( {2{m^2} + 1} \right) = 0\\
\to 1 - 8m + 16{m^2} - 16{m^2} - 8 = 0\\
\to - 7 = 8m\\
\to m = - \dfrac{7}{8}\\
c)Ycbt \Leftrightarrow 1 - 8m + 16{m^2} - 8\left( {2{m^2} + 1} \right) > 0\\
\to 1 - 8m + 16{m^2} - 16{m^2} - 8 > 0\\
\to - 7 > 8m\\
\to m < - \dfrac{7}{8}\\
d)4\left( {{x_1} + {x_2}} \right) + 2{x_1}{x_2} - 5 = 0\\
\to 4.\dfrac{{4m - 1}}{2} + 2.\dfrac{{2{m^2} + 1}}{2} - 5 = 0\\
\to 8m - 2 + 2{m^2} + 1 - 5 = 0\\
\to 2{m^2} + 8m - 6 = 0\\
\to \left[ \begin{array}{l}
m = - 2 + \sqrt 7 \left( l \right)\\
m = - 2 - \sqrt 7 \left( {TM} \right)
\end{array} \right.\\
e)4\left( {{x_1}^2 + {x_2}^2} \right) - 21 = 0\\
\to 4\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2}} \right) - 21 = 0\\
\to 4{\left( {{x_1} + {x_2}} \right)^2} - 8{x_1}{x_2} - 21 = 0\\
\to 4.{\left( {\dfrac{{4m - 1}}{2}} \right)^3} - 8.\dfrac{{2{m^2} + 1}}{2} - 21 = 0\\
\to 16{m^2} - 8m + 1 - 8{m^2} - 4 - 21 = 0\\
\to 8{m^2} - 8m - 20 = 0\\
\to \left[ \begin{array}{l}
m = \dfrac{{1 + \sqrt {11} }}{2}\left( l \right)\\
m = \dfrac{{1 - \sqrt {11} }}{2}\left( {TM} \right)
\end{array} \right.
\end{array}\)
