Giải thích các bước giải:
Câu 2:
\(\begin{array}{l}
a.B = \left[ {\frac{{2 - \sqrt x + x}}{{1 - \sqrt x }}} \right]:\left[ {\frac{{1 - \sqrt x + 2\sqrt x }}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}} \right]\\
= \frac{{2 - \sqrt x + x}}{{1 - \sqrt x }}.\frac{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}{{1 + \sqrt x }}\\
= 2 - \sqrt x + x\\
b.B = 5\\
\to 2 - \sqrt x + x = 5\\
\to - 3 - \sqrt x + x = 0\\
\to \left[ \begin{array}{l}
\sqrt x = \frac{{1 + \sqrt {13} }}{2}\\
\sqrt x = \frac{{1 - \sqrt {13} }}{2}\left( l \right)
\end{array} \right.\\
\to x = \frac{{7 + \sqrt {13} }}{2}\left( {TM} \right)
\end{array}\)
Câu 1:
\(\begin{array}{l}
a.ĐK:x \ge 0;x \ne 1\\
A = \frac{{x + \sqrt x - 2\sqrt x + 2 - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \frac{{x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \frac{{\sqrt x }}{{\sqrt x + 1}}\\
b.x = 9 \to A = \frac{{\sqrt 9 }}{{\sqrt 9 + 1}} = \frac{3}{4}\\
c.\frac{{\sqrt x }}{{\sqrt x + 1}} = \frac{{\sqrt x + 1 - 1}}{{\sqrt x + 1}} = 1 - \frac{1}{{\sqrt x + 1}}
\end{array}\)
Để A nguyên
⇔ \(\frac{1}{{\sqrt x + 1}}\) nguyên
\(\begin{array}{l}
\Leftrightarrow \sqrt x + 1 \in U\left( 1 \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x + 1 = 1\\
\sqrt x + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 0\left( {TM} \right)\\
\sqrt x = - 2\left( l \right)
\end{array} \right.\\
\to x = 0
\end{array}\)
\(\begin{array}{l}
C4:\\
P = \left( {\frac{{\sqrt x + 1}}{{\sqrt x }}} \right).\left[ {\frac{{\sqrt x + 1 + \sqrt x - 1 - 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right]\\
= \left( {\frac{{\sqrt x + 1}}{{\sqrt x }}} \right).\frac{{2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{2}{{\sqrt x }}\\
P > 1\\
\to \frac{2}{{\sqrt x }} > 1\\
\to \frac{{2 - \sqrt x }}{{\sqrt x }} > 0\\
\to \left\{ \begin{array}{l}
\sqrt x > 0\left( {ld\forall x > 0} \right)\\
2 - \sqrt x > 0
\end{array} \right.\\
\to 4 > x > 1
\end{array}\)
