Đáp án:
B3:
b. \(0 \le x < 9\)
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
a.DK:x \ge 0;x \ne 1\\
b.P = \left[ {\dfrac{{3 + \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right].\left( {\sqrt x + 1} \right)\\
= \dfrac{{2 + \sqrt x }}{{\sqrt x - 1}}\\
P = \dfrac{5}{4}\\
\to \dfrac{{2 + \sqrt x }}{{\sqrt x - 1}} = \dfrac{5}{4}\\
\to 8 + 4\sqrt x = 5\sqrt x - 5\\
\to \sqrt x = 13\\
\to x = 169\\
B3:\\
a.DK:x \ge 0;x \ne 9\\
b.D = \left[ {\dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}} \right]:\left( {\dfrac{{2\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}}} \right)\\
= \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
= \dfrac{{ - 3\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3}}{{\sqrt x + 3}}\\
D < - \dfrac{1}{2}\\
\to \dfrac{{ - 3}}{{\sqrt x + 3}} < - \dfrac{1}{2}\\
\to \dfrac{3}{{\sqrt x + 3}} > \dfrac{1}{2}\\
\to \dfrac{{6 - \sqrt x - 3}}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to 3 - \sqrt x > 0\left( {do:\sqrt x + 3 > 0\forall x \ge 0} \right)\\
\to 9 > x\\
\to 0 \le x < 9
\end{array}\)
