Đáp án:
\(\begin{array}{l}
B2:\\
a)x - \sqrt x + 2\\
b)x = \dfrac{{3 + \sqrt 5 }}{2}\\
c)Min = \dfrac{7}{4}\\
d)\left[ \begin{array}{l}
x = 9\\
x = 4
\end{array} \right.\\
e)x \in \emptyset \\
B3:\\
a)\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
b)x \in \emptyset \\
c)\dfrac{1}{4} > x > 0
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
a)P = \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{x + \sqrt x + 1}} - 2\sqrt x - 1 + \dfrac{{2\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x - 1}}\\
= x - \sqrt x - 2\sqrt x - 1 + 2\sqrt x + 2\\
= x - \sqrt x + 2\\
b)P = 3\\
\to x - \sqrt x + 2 = 3\\
\to x - \sqrt x - 1 = 0\\
\to \left[ \begin{array}{l}
\sqrt x = \dfrac{{1 + \sqrt 5 }}{2}\\
\sqrt x = \dfrac{{1 - \sqrt 5 }}{2}\left( l \right)
\end{array} \right.\\
\to x = \dfrac{{3 + \sqrt 5 }}{2}\\
c)P = x - \sqrt x + 2\\
= x - 2\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{7}{4}\\
= {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{7}{4}\\
Do:{\left( {\sqrt x - \dfrac{1}{2}} \right)^2} \ge 0\forall x > 0\\
\to {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{7}{4} \ge \dfrac{7}{4}\\
\to Min = \dfrac{7}{4}\\
\Leftrightarrow x = \dfrac{1}{4}\\
d)Q = \dfrac{P}{{\sqrt x - 1}} = \dfrac{{x - \sqrt x + 2}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right) + 2}}{{\sqrt x - 1}}\\
= \sqrt x + \dfrac{2}{{\sqrt x - 1}}\\
Q \in Z \to \dfrac{2}{{\sqrt x - 1}} \in Z\\
\to \sqrt x - 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 1 = 2\\
\sqrt x - 1 = 1\\
\sqrt x - 1 = - 1\\
\sqrt x - 1 = - 2\left( l \right)
\end{array} \right. \to \left[ \begin{array}{l}
x = 9\\
x = 4\\
x = 0\left( l \right)
\end{array} \right.\\
e)P < 1\\
\to x - \sqrt x + 2 < 1\\
\to x - \sqrt x + 1 < 0\left( {vô lý} \right)\\
\to x \in \emptyset \\
B3:\\
a)P = \dfrac{{x - 2 + \sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
b)P = \dfrac{9}{{10}}\\
\to \dfrac{{\sqrt x + 1}}{{\sqrt x }} = \dfrac{9}{{10}}\\
\to 10\sqrt x + 10 = 9\sqrt x \\
\to \sqrt x = - 10\left( l \right)\\
\to x \in \emptyset \\
c)P > 3\\
\to \dfrac{{\sqrt x + 1}}{{\sqrt x }} > 3\\
\to \dfrac{{\sqrt x + 1 - 3\sqrt x }}{{\sqrt x }} > 0\\
\to 1 - 2\sqrt x > 0\left( {do:\sqrt x > 0\forall x > 0} \right)\\
\to \dfrac{1}{4} > x > 0
\end{array}\)
