Đáp án:
5) \(\left[ \begin{array}{l}
x = 64\\
x = 4\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ge 0;x \ne 9\\
P = \dfrac{{x\sqrt x - 3}}{{x - 2\sqrt x - 3}} - \dfrac{{2\left( {\sqrt x - 3} \right)}}{{\sqrt x + 1}} + \dfrac{{\sqrt x + 3}}{{3 - \sqrt x }}\\
= \dfrac{{x\sqrt x - 3 - 2{{\left( {\sqrt x - 3} \right)}^2} - \left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x\sqrt x - 3 - 2\left( {x - 6\sqrt x + 9} \right) - x - 4\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x\sqrt x - x - 4\sqrt x - 6 - 2x + 12\sqrt x - 18}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x\sqrt x - 3x + 8\sqrt x - 24}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 3} \right)\left( {x + 8} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 8}}{{\sqrt x + 1}}\\
2)Thay:x = 14 - 6\sqrt 5 \\
= 9 - 2.3.\sqrt 5 + 5\\
= {\left( {3 - \sqrt 5 } \right)^2}\\
\to P = \dfrac{{14 - 6\sqrt 5 + 8}}{{\sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}} + 1}} = \dfrac{{22 - 6\sqrt 5 }}{{3 - \sqrt 5 + 1}}\\
= \dfrac{{22 - 6\sqrt 5 }}{{4 - \sqrt 5 }}\\
3)P = 8\\
\to \dfrac{{x + 8}}{{\sqrt x + 1}} = 8\\
\to x + 8 = 8\sqrt x + 8\\
\to x - 8\sqrt x = 0\\
\to \sqrt x \left( {\sqrt x - 8} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 64
\end{array} \right.\\
5)P = \dfrac{{x + 8}}{{\sqrt x + 1}} = \dfrac{{x - 1 + 9}}{{\sqrt x + 1}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + 9}}{{\sqrt x + 1}}\\
= \left( {\sqrt x - 1} \right) + \dfrac{9}{{\sqrt x + 1}}\\
P \in Z \to \dfrac{9}{{\sqrt x + 1}} \in Z\\
\to \sqrt x + 1 \in U\left( 9 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 1 = 9\\
\sqrt x + 1 = 3\\
\sqrt x + 1 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 64\\
x = 4\\
x = 0
\end{array} \right.
\end{array}\)
