Đáp án:
\(\begin{array}{l}
a){a^2} + a\\
b) - \dfrac{1}{{2a\sqrt 2 }}\\
c)\dfrac{{4x - 5}}{{x - 3}}\\
d)\left[ \begin{array}{l}
\dfrac{{\sqrt x - 2}}{{\sqrt x + 2}}\\
- \dfrac{{\sqrt x - 2}}{{\sqrt x + 2}}
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\sqrt {{a^2}{{\left( {a + 1} \right)}^2}} = \left| {a\left( {a + 1} \right)} \right|\\
= a\left( {a + 1} \right)\\
= {a^2} + a\\
b)\dfrac{{\sqrt {16{a^4}{b^6}} }}{{\sqrt {128{a^6}{b^6}} }} = \sqrt {\dfrac{1}{{8{a^2}}}} \\
= \dfrac{1}{{2\left| a \right|\sqrt 2 }} = - \dfrac{1}{{2a\sqrt 2 }}\\
c)\sqrt {\dfrac{{{{\left( {x - 2} \right)}^4}}}{{{{\left( {x - 3} \right)}^2}}}} + \dfrac{{{x^2} - 1}}{{x - 3}}\\
= - \dfrac{{{{\left( {x - 2} \right)}^2}}}{{x - 3}} + \dfrac{{{x^2} - 1}}{{x - 3}}\\
= \dfrac{{ - {x^2} + 4x - 4 + {x^2} - 1}}{{x - 3}}\\
= \dfrac{{4x - 5}}{{x - 3}}\\
d)DK:x \ge 0\\
\sqrt {\dfrac{{x - 4\sqrt x + 4}}{{x + 4\sqrt x + 4}}} = \sqrt {\dfrac{{{{\left( {\sqrt x - 2} \right)}^2}}}{{{{\left( {\sqrt x + 2} \right)}^2}}}} \\
= \left| {\dfrac{{\sqrt x - 2}}{{\sqrt x + 2}}} \right| = \left[ \begin{array}{l}
\dfrac{{\sqrt x - 2}}{{\sqrt x + 2}}\left( {x \ge 4} \right)\\
- \dfrac{{\sqrt x - 2}}{{\sqrt x + 2}}\left( {0 \le x < 4} \right)
\end{array} \right.
\end{array}\)
