Đáp án:
d. \(\left[ \begin{array}{l}
m = \dfrac{1}{2}\\
m = - \dfrac{1}{2}
\end{array} \right.\)
Giải thích các bước giải:
a. Để phương trình có nghiệm
\(\begin{array}{l}
\to {m^2} + 2m + 1 - 4m \ge 0\\
\to {m^2} - 2m + 1 \ge 0\\
\to {\left( {m - 1} \right)^2} \ge 0\left( {ld} \right)\forall m \in R\\
b.Thay:m = 2\\
Pt \to {x^2} - 6x + 8 = 0\\
\to {x^2} - 2x - 4x + 8 = 0\\
\to x\left( {x - 2} \right) - 4\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x - 4 = 0\\
x - 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
x = 2
\end{array} \right.\\
c.Thay:x = - 2\\
\to 4 + 4\left( {m + 1} \right) + 4m = 0\\
\to 8m + 8 = 0\\
\to m = - 1\\
Thay:m = - 1\\
Pt \to {x^2} - 4 = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 2
\end{array} \right.\\
d.{x_1}^2 + {x_2}^2 = 5\\
\to {x_1}^2 + {x_2}^2 + 2{x_1}{x_2} - 2{x_1}{x_2} = 5\\
\to {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 5\\
\to {\left( {2m + 2} \right)^2} - 2.4m = 5\\
\to 4{m^2} + 8m + 4 - 8m = 5\\
\to 4{m^2} = 1\\
\to \left[ \begin{array}{l}
m = \dfrac{1}{2}\\
m = - \dfrac{1}{2}
\end{array} \right.
\end{array}\)
