Đáp án:
$\begin{array}{l}
1)a)m = 1\\
\Leftrightarrow {x^2} - 2\left( {m - 1} \right).x + {m^2} - 6 = 0\\
Thay\,m = 1\\
\Leftrightarrow {x^2} + 1 - 6 = 0\\
\Leftrightarrow {x^2} = 5\\
\Leftrightarrow x = \pm \sqrt 5 \\
b)\Delta ' > 0\\
\Leftrightarrow {\left( {m - 1} \right)^2} - {m^2} + 6 > 0\\
\Leftrightarrow {m^2} - 2m + 1 - {m^2} + 6 > 0\\
\Leftrightarrow 2m < 7\\
\Leftrightarrow m < \dfrac{7}{2}\\
Vay\,m < \dfrac{7}{2}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m - 1} \right)\\
{x_1}{x_2} = {m^2} - 6
\end{array} \right.\\
P = x_1^2 + x_2^2 - {x_1}{x_2}\\
= {\left( {{x_1} + {x_2}} \right)^2} - 3{x_1}{x_2}\\
= 4{\left( {m - 1} \right)^2} - 3.\left( {{m^2} - 6} \right)\\
= 4{m^2} - 8m + 4 - 3{m^2} + 18\\
= {m^2} - 8m + 16 + 6\\
= {\left( {m - 4} \right)^2} + 6 \ge 6\\
\Leftrightarrow P \ge 6\\
\Leftrightarrow GTNN:P = 6\,khi:m = 4\left( {ktmdk} \right)
\end{array}$
Vậy ko có giá trị của m để P đạt GTNN
$\begin{array}{l}
a)Dkxd:x \ge 0;x\# 4\\
A = \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x + 2}} - \dfrac{{2 + 5\sqrt x }}{{x - 4}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) + 2\sqrt x \left( {\sqrt x - 2} \right) - 2 - 5\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + 3\sqrt x + 2 + 2x - 4\sqrt x - 2 - 5\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3x - 6\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 2}}\\
b)x = 9 - 4\sqrt 5 \left( {tmdk} \right)\\
\Leftrightarrow x = {\left( {\sqrt 5 - 2} \right)^2}\\
\Leftrightarrow \sqrt x = \sqrt 5 - 2\\
\Leftrightarrow A = \dfrac{{3\sqrt x }}{{\sqrt x + 2}} = \dfrac{{3\sqrt 5 - 6}}{{\sqrt 5 }} = \dfrac{{15 - 6\sqrt 5 }}{5}
\end{array}$
