Đáp án:
\(\begin{array}{l}
a)\dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
b)0 < x < 4\\
c)x = 1\\
d)\dfrac{7}{8}\\
e)x \in \emptyset \\
f)x = - 4\left( {KTM} \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x > 0;x \ne 4\\
P = \dfrac{{\sqrt x + 1 - \sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
b)P < \dfrac{1}{2}\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x }} < \dfrac{1}{2}\\
\to \dfrac{{2\sqrt x - 2 - \sqrt x }}{{2\sqrt x }} < 0\\
\to \dfrac{{\sqrt x - 2}}{{2\sqrt x }} < 0\\
\to \sqrt x - 2 < 0\left( {do:\sqrt x > 0} \right)\\
\to 0 < x < 4\\
c)P = \dfrac{{\sqrt x - 1}}{{\sqrt x }} = 1 - \dfrac{1}{{\sqrt x }}\\
P \in Z \to \dfrac{1}{{\sqrt x }} \in Z\\
\to \sqrt x \in U\left( 1 \right)\\
\to x = 1\\
d)Thay:x = 64\\
\to P = \dfrac{{\sqrt {64} - 1}}{{\sqrt {64} }} = \dfrac{7}{8}\\
e)P = 4\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x }} = 4\\
\to \sqrt x - 1 = 4\sqrt x \\
\to 3\sqrt x = - 1\left( {KTM} \right)\\
\to x \in \emptyset \\
f)Thay:x = \sqrt {9 - 4\sqrt 5 } - \sqrt {9 + 4\sqrt 5 } \\
= \sqrt {5 - 2.2\sqrt 5 + 5} - \sqrt {5 + 2.2\sqrt 5 + 5} \\
= \sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} - \sqrt {{{\left( {\sqrt 5 + 2} \right)}^2}} \\
= \sqrt 5 - 2 - \sqrt 5 - 2 = - 4\left( {KTM} \right)
\end{array}\)
