Đáp án:
B3:
a) \(\dfrac{{2\sqrt x }}{{3\left( {\sqrt x + 3} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
a)DK:x \ge 0;x \ne 9\\
A = \left[ {\dfrac{{\sqrt x + 3 + \sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}} \right].\dfrac{{\sqrt x - 3}}{3}\\
= \dfrac{{2\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{3}\\
= \dfrac{{2\sqrt x }}{{3\left( {\sqrt x + 3} \right)}}\\
b)A > \dfrac{1}{3}\\
\to \dfrac{{2\sqrt x }}{{3\left( {\sqrt x + 3} \right)}} > \dfrac{1}{3}\\
\to \dfrac{{2\sqrt x }}{{\sqrt x + 3}} > 1\\
\to \dfrac{{2\sqrt x - \sqrt x - 3}}{{\sqrt x + 3}} > 0\\
\to \sqrt x - 3 > 0\left( {do:\sqrt x + 3 > 0\forall x \ge 0} \right)\\
\to x > 9\\
B4:\\
a)DK:x \ge 0;x \ne 1\\
P = \left[ {\dfrac{{ - 3 + \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right].\left( {\sqrt x + 1} \right)\\
= \dfrac{{\sqrt x - 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\left( {\sqrt x + 1} \right)\\
= \dfrac{{\sqrt x - 4}}{{\sqrt x - 1}}\\
b)P = \dfrac{5}{4}\\
\to \dfrac{{\sqrt x - 4}}{{\sqrt x - 1}} = \dfrac{5}{4}\\
\to 4\sqrt x - 16 = 5\sqrt x - 5\\
\to \sqrt x = - 9\left( l \right)\\
\to x \in \emptyset
\end{array}\)
