Đáp án:
3) \(0 < x \le 9;x \ne \left\{ {1;4} \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:x = 7 + 4\sqrt 3 = 4 + 2.2.\sqrt 3 + 3\\
= {\left( {2 + \sqrt 3 } \right)^2}\\
\to A = \dfrac{{\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} - 1}}{{\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} - 2}}\\
= \dfrac{{2 + \sqrt 3 - 1}}{{2 + \sqrt 3 - 2}} = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 }}\\
2)B = \dfrac{{x - \sqrt x + 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}} - \dfrac{x}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x\sqrt x - x + 2\sqrt x - x\sqrt x - x}}{{\sqrt x \left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 2x + 2\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 2\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - 2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}\\
P = B:A = \dfrac{{ - 2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{{\sqrt x - 1}}{{\sqrt x - 2}}\\
= \dfrac{{ - 2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x - 2}}{{\sqrt x - 1}}\\
= \dfrac{{ - 2}}{{\sqrt x + 1}}\\
3)P\sqrt x \ge - \dfrac{3}{2}\\
\to \dfrac{{ - 2\sqrt x }}{{\sqrt x + 1}} \ge - \dfrac{3}{2}\\
\to \dfrac{{2\sqrt x }}{{\sqrt x + 1}} \le \dfrac{3}{2}\\
\to \dfrac{{4\sqrt x - 3\sqrt x - 3}}{{2\left( {\sqrt x + 1} \right)}} \le 0\\
\to \dfrac{{\sqrt x - 3}}{{2\left( {\sqrt x + 1} \right)}} \le 0\\
Do:2\left( {\sqrt x + 1} \right) > 0\forall x > 0\\
\to \sqrt x - 3 \le 0\\
\to \sqrt x \le 3\\
\to x \le 9\\
KL:0 < x \le 9;x \ne \left\{ {1;4} \right\}
\end{array}\)
