Đáp án:
a) \(\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \dfrac{{2\sqrt x - 9 + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right) - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{2\sqrt x - 9 + 2x - 4\sqrt x + \sqrt x - 2 - x + 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
b)Thay:x = 3 + 2\sqrt 2 \\
= 2 + 2\sqrt 2 .1 + 1 = {\left( {\sqrt 2 + 1} \right)^2}\\
\to P = \dfrac{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + 1}}{{\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - 3}}\\
= \dfrac{{\sqrt 2 + 1 + 1}}{{\sqrt 2 + 1 - 3}}\\
= \dfrac{{\sqrt 2 + 2}}{{\sqrt 2 - 2}} = - 3 - 2\sqrt 2 \\
c)P = - 3\\
\to \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = - 3\\
\to \sqrt x + 1 = - 3\sqrt x + 9\\
\to 4\sqrt x = 8\\
\to \sqrt x = 2\\
\to x = 4\left( l \right)\\
\to x \in \emptyset
\end{array}\)
