Đáp án:
\(\begin{array}{l}
n)\dfrac{1}{{\sqrt x + 2}}\\
o)\dfrac{{4\sqrt x }}{{\sqrt x - 3}}\\
p)\left[ \begin{array}{l}
P = 2\sqrt {m - 1} \\
P = 2
\end{array} \right.\\
q)\left[ \begin{array}{l}
Q = 2\sqrt {m + 1} \\
Q = 2
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
n)N = \dfrac{{2\sqrt x + x - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{\sqrt x + 2}}{{x + \sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x + 2}}\\
= \dfrac{1}{{\sqrt x + 2}}\\
o)O = \dfrac{{{{\left( {2 + \sqrt x } \right)}^2} - {{\left( {2 - \sqrt x } \right)}^2} + 4x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}:\dfrac{{{{\left( {\sqrt x - 3} \right)}^2}}}{{\left( {2 - \sqrt x } \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{4 + 4\sqrt x + x - 4 + 4\sqrt x - x + 4x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\left( {2 - \sqrt x } \right)\left( {\sqrt x - 3} \right)}}{{{{\left( {\sqrt x - 3} \right)}^2}}}\\
= \dfrac{{4x + 8\sqrt x }}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\left( {2 - \sqrt x } \right)\left( {\sqrt x - 3} \right)}}{{{{\left( {\sqrt x - 3} \right)}^2}}}\\
= \dfrac{{4\sqrt x \left( {2 + \sqrt x } \right)}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{\left( {2 - \sqrt x } \right)\left( {\sqrt x - 3} \right)}}{{{{\left( {\sqrt x - 3} \right)}^2}}}\\
= \dfrac{{4\sqrt x }}{{\sqrt x - 3}}\\
p)P = \sqrt {m - 1 + 2\sqrt {m - 1} .1 + 1} + \sqrt {m - 1 - 2\sqrt {m - 1} .1 + 1} \\
= \sqrt {{{\left( {\sqrt {m - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {m - 1} - 1} \right)}^2}} \\
= \left| {\sqrt {m - 1} + 1} \right| + \left| {\sqrt {m - 1} - 1} \right|\\
\to \left[ \begin{array}{l}
P = \sqrt {m - 1} + 1 + \sqrt {m - 1} - 1\left( {DK:\sqrt {m - 1} \ge 1} \right)\\
P = \sqrt {m - 1} + 1 - \sqrt {m - 1} + 1\left( {DK:\sqrt {m - 1} < 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
P = 2\sqrt {m - 1} \\
P = 2
\end{array} \right.\\
q)Q = \sqrt {m + 1 + 2\sqrt {m + 1} .1 + 1} + \sqrt {m + 1 - 2\sqrt {m + 1} .1 + 1} \\
= \sqrt {{{\left( {\sqrt {m + 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {m + 1} - 1} \right)}^2}} \\
= \left| {\sqrt {m + 1} + 1} \right| + \left| {\sqrt {m + 1} - 1} \right|\\
\to \left[ \begin{array}{l}
Q = \sqrt {m + 1} + 1 + \sqrt {m + 1} - 1\left( {DK:\sqrt {m + 1} \ge 1} \right)\\
Q = \sqrt {m + 1} + 1 - \sqrt {m + 1} + 1\left( {DK:\sqrt {m + 1} < 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
Q = 2\sqrt {m + 1} \\
Q = 2
\end{array} \right.
\end{array}\)
