Đáp án:
b. \(\left[ \begin{array}{l}
m = \frac{3}{2}\\
m = \frac{3}{4}
\end{array} \right.\)
Giải thích các bước giải:
a. Xét:
\(\begin{array}{l}
Δ'= {m^2} - 2m + 1\\
= {\left( {m - 1} \right)^2} \ge 0\forall m \in R\\
\to dpcm\\
b. Δ'= {\left( {m - 1} \right)^2}\\
\to \left[ \begin{array}{l}
x = m + \sqrt {{{\left( {m - 1} \right)}^2}} \\
x = m - \sqrt {{{\left( {m - 1} \right)}^2}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2m - 1\\
x = 1
\end{array} \right.\\
Có:{x_1} = 2{x_2}\\
\to \left[ \begin{array}{l}
2m - 1 = 2\\
1 = 2\left( {2m - 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = \frac{3}{2}\\
1 = 4m - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = \frac{3}{2}\\
m = \frac{3}{4}
\end{array} \right.
\end{array}\)
