Đáp án:
2) \(\dfrac{{\sqrt x + 6}}{{\sqrt x - 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)Thay:x = 4 + 2\sqrt 3 = 3 + 2\sqrt 3 .1 + 1\\
= {\left( {\sqrt 3 + 1} \right)^2}\\
\to A = \dfrac{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - 1}}{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} + 2}} = \dfrac{{\sqrt 3 + 1 - 1}}{{\sqrt 3 + 1 + 2}} = \dfrac{{\sqrt 3 }}{{\sqrt 3 + 3}}\\
b)B = \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right) + 5\left( {\sqrt x + 1} \right) + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x - 3 + 5\sqrt x + 5 + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 7\sqrt x + 6}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \dfrac{{\left( {\sqrt x + 6} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 6}}{{\sqrt x - 1}}\\
3)P = A.B = \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}}.\dfrac{{\sqrt x + 6}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x + 6}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 + 4}}{{\sqrt x + 2}} = 1 + \dfrac{4}{{\sqrt x + 2}}\\
P \in Z \Leftrightarrow \dfrac{4}{{\sqrt x + 2}} \in Z\\
\to \left[ \begin{array}{l}
\sqrt x + 2 = 4\\
\sqrt x + 2 = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
x = 0
\end{array} \right.
\end{array}\)
