Đáp án:
2) m>2013
Giải thích các bước giải:
\(\begin{array}{l}
1)A = \dfrac{{2\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} + \dfrac{{\sqrt x + 1}}{{\sqrt x + 3}} + 4\\
= \dfrac{2}{{\sqrt x + 3}} + \dfrac{{\sqrt x + 1}}{{\sqrt x + 3}} + 4\\
= \dfrac{{\sqrt x + 3 + 4\sqrt x + 12}}{{\sqrt x + 3}}\\
= \dfrac{{5\sqrt x + 15}}{{\sqrt x + 3}} = \dfrac{{5\left( {\sqrt x + 3} \right)}}{{\sqrt x + 3}} = 5
\end{array}\)
2) Để phương trình có nghiệm
\(\begin{array}{l}
\to \Delta \ge 0\\
\to {m^2} - 4\left( {m - 1} \right) \ge 0\\
\to {m^2} - 4m + 4 \ge 0\\
\to {\left( {m - 2} \right)^2} \ge 0\left( {ld} \right)\forall m\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = m\\
{x_1}{x_2} = m - 1
\end{array} \right.\\
TH1:{x_2} > 2012;2012 = {x_1}\\
Thay:x = 2012\\
Pt \to {2012^2} - 2012m + m - 1 = 0\\
\to m = 2013\\
Thay:m = 2013\\
Pt \to {x^2} - 2013x + 2012 = 0\\
\to \left( {x - 2012} \right)\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2012\\
x = 1
\end{array} \right.\left( {KTM} \right)\\
\to m = 2013\left( l \right)\\
TH2:{x_2} > 2012 > {x_1}\\
\to \left\{ \begin{array}{l}
{x_2} - 2012 > 0\\
{x_1} - 2012 < 0
\end{array} \right.\\
\to \left( {{x_1} - 2012} \right)\left( {{x_2} - 2012} \right) < 0\\
\to {x_1}{x_2} - 2012\left( {{x_1} + {x_2}} \right) + {2012^2} < 0\\
\to m - 1 - 2012m + {2012^2} < 0\\
\to 2013 < m
\end{array}\)
