Đáp án:
C3:
b) \(\left[ \begin{array}{l}
x = 1 + \sqrt 5 \\
x = 1 - \sqrt 5
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
C3:\\
a)DK:\left\{ \begin{array}{l}
m \ne 0\\
4{\left( {m - 1} \right)^2} - m.\left( { - 8} \right) = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 0\\
4{m^2} - 8m + 4 + 8m = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 0\\
4{m^2} + 4 = 0\left( {vô lý} \right)
\end{array} \right.\\
\to m \in \emptyset \\
b)Thay:m = 2\\
Pt \to 2{x^2} - 4x - 8 = 0\\
\to {x^2} - 2x - 4 = 0\\
\Delta ' = 1 + 4 = 5\\
\to \left[ \begin{array}{l}
x = 1 + \sqrt 5 \\
x = 1 - \sqrt 5
\end{array} \right.\\
C4:\\
Xét:\Delta ' = {m^2} + 2m + 1 - 4m + {m^2}\\
= 2{m^2} - 2m + 1\\
= 2{m^2} - 2.m\sqrt 2 .\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} + \dfrac{1}{2}\\
= {\left( {m\sqrt 2 - \dfrac{1}{{\sqrt 2 }}} \right)^2} + \dfrac{1}{2} > 0\forall m\\
\to dpcm
\end{array}\)
