Đáp án:
$\begin{array}{l}
e)A = \sqrt {\dfrac{{{{\left( {x - 2} \right)}^4}}}{{{{\left( {3 - x} \right)}^2}}}} + \dfrac{{{x^2} - 1}}{{x - 3}}\\
= \dfrac{{{{\left( {x - 2} \right)}^2}}}{{3 - x}} + \dfrac{{{x^2} - 1}}{{x - 3}}\left( {do:x < 3} \right)\\
= \dfrac{{{x^2} - 4x + 4 - {x^2} + 1}}{{3 - x}}\\
= \dfrac{{ - 4x + 5}}{{3 - x}}\\
Khi:x = 0,5\\
\Leftrightarrow A = \dfrac{{ - 4x + 5}}{{3 - x}} = \dfrac{{ - 4.0,5 + 5}}{{3 - 0,5}} = \dfrac{3}{{2,5}} = \dfrac{6}{5}\\
B6)\\
a)\sqrt {{x^2} - 4} + 2\sqrt {x - 2} \\
Dkxd:\left\{ \begin{array}{l}
{x^2} - 4 \ge 0\\
x - 2 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} \ge 4\\
x \ge 2
\end{array} \right. \Leftrightarrow x \ge 2\\
Vậy\,x \ge 2\\
b)3\sqrt {x + 3} + \sqrt {{x^2} - 9} \\
Dkxd:\left\{ \begin{array}{l}
x + 3 \ge 0\\
{x^2} - 9 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 3\\
{x^2} \ge 9
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 3\\
\left[ \begin{array}{l}
x \ge 3\\
x \le - 3
\end{array} \right.
\end{array} \right. \Leftrightarrow x \ge 3\\
Vậy\,x \ge 3
\end{array}$
