Đáp án:
a) \({\dfrac{{ - \sqrt x + 2}}{{x - 1}}}\)
Giải thích các bước giải:
\(\begin{array}{l}
\begin{array}{*{20}{l}}
{a)DK:x \ge 0;x \ne 1}\\
{A = \dfrac{{2\sqrt x + 2 - 2\sqrt x + 2 - 2\sqrt x }}{{2\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}}\\
{ = \dfrac{{ - 2\sqrt x + 4}}{{2\left( {x - 1} \right)}}}\\
{ = \dfrac{{ - \sqrt x + 2}}{{x - 1}}}\\
{b)Thay:x = \dfrac{4}{9}}\\
{ \to \sqrt x = \dfrac{2}{3}}\\
{ \to A = \dfrac{{ - \dfrac{2}{3} + 2}}{{\dfrac{4}{9} - 1}} = \dfrac{4}{3}:\left( { - \dfrac{5}{9}} \right) = - \dfrac{{12}}{5}}\\
{c)A > \dfrac{1}{2}}\\
{ \to \dfrac{{ - \sqrt x + 2}}{{x - 1}} > \dfrac{1}{2}}\\
{ \to \dfrac{{ - 2\sqrt x + 4 - x + 1}}{{2\left( {x - 1} \right)}} > 0}\\
{ \to \dfrac{{ - x - 2\sqrt x + 5}}{{2\left( {x - 1} \right)}} > 0}\\
{TH1:x - 1 > 0 \to x > 1}\\
{ \to - x - 2\sqrt x + 5 > 0}\\
\begin{array}{l}
\to 0 \le \sqrt x < - 1 + \sqrt 6 \\
\to 0 \le x < 7 - 2\sqrt 6
\end{array}
\end{array}\\
\to 1 < x < 7 - 2\sqrt 6 \\
TH2:x - 1 < 0 \to 0 \le x < 1\\
\to - x - 2\sqrt x + 5 < 0\\
\to \sqrt x > - 1 + \sqrt 6 \\
\to x > 7 - 2\sqrt 6 \\
KL:\left[ \begin{array}{l}
x > 7 - 2\sqrt 6 \\
1 < x < 7 - 2\sqrt 6
\end{array} \right.
\end{array}\)
