Đáp án:
\(\begin{array}{l}
a)\dfrac{{2x + 3\sqrt x - 10}}{{x - 5\sqrt x + 6}}\\
b)x = 48 - 32\sqrt 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 9;x \ne 4\\
P = \left( {1 - \dfrac{{1 - x - 3\sqrt x }}{{x - 9}}} \right):\left( {\dfrac{{\sqrt x - 3}}{{2 - \sqrt x }} + \dfrac{{\sqrt x - 2}}{{\sqrt x + 3}} - \dfrac{{9 - x}}{{x + \sqrt x - 6}}} \right)\\
= \dfrac{{x - 9 - 1 + x + 3\sqrt x }}{{x - 9}}:\dfrac{{ - \left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right) + {{\left( {\sqrt x - 2} \right)}^2} - 9 + x}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x + 3\sqrt x - 10}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}{{ - x + 9 + {{\left( {\sqrt x - 2} \right)}^2} - 9 + x}}\\
= \dfrac{{2x + 3\sqrt x - 10}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 3} \right)}}{{{{\left( {\sqrt x - 2} \right)}^2}}}\\
= \dfrac{{2x + 3\sqrt x - 10}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2x + 3\sqrt x - 10}}{{x - 5\sqrt x + 6}}\\
b)P = 1\\
\to \dfrac{{2x + 3\sqrt x - 10}}{{x - 5\sqrt x + 6}} = 1\\
\to 2x + 3\sqrt x - 10 = x - 5\sqrt x + 6\\
\to x + 8\sqrt x - 16 = 0\\
\to \left[ \begin{array}{l}
\sqrt x = - 4 + 4\sqrt 2 \\
\sqrt x = - 4 - 4\sqrt 2 \left( l \right)
\end{array} \right.\\
\to x = 48 - 32\sqrt 2
\end{array}\)
( đề của b thiếu dấu ngoặc nên t đã sửa thêm vô r nhé )
