Đáp án:
\(\begin{array}{l}
a)P = \dfrac{3}{{\sqrt x + 3}}\\
b)\dfrac{3}{{\sqrt {11} }}\\
c)9 > x \ge 0
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \dfrac{{3x + 3 - 2\sqrt x \left( {\sqrt x - 3} \right) - \sqrt x \left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{{2\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}}\\
= \dfrac{{3x + 3 - 2x + 6\sqrt x - x - 3\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{3\sqrt x + 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{3}{{\sqrt x + 3}}\\
b)Thay:x = 20 - 6\sqrt {11} \\
= 11 - 2.3.\sqrt {11} + 9\\
= {\left( {\sqrt {11} - 3} \right)^2}\\
\to P = \dfrac{3}{{\sqrt {{{\left( {\sqrt {11} - 3} \right)}^2}} + 3}} = \dfrac{3}{{\sqrt {11} - 3 + 3}}\\
= \dfrac{3}{{\sqrt {11} }}\\
c)P > \dfrac{1}{2}\\
\to \dfrac{3}{{\sqrt x + 3}} > \dfrac{1}{2}\\
\to \dfrac{{6 - \sqrt x - 3}}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to \dfrac{{3 - \sqrt x }}{{2\left( {\sqrt x + 3} \right)}} > 0\\
\to 3 - \sqrt x > 0\left( {do:\sqrt x + 3 > 0\forall x} \right)\\
\to 9 > x \ge 0
\end{array}\)
( câu d thiếu đề nha bạn )
