Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{\left( {3 - a} \right)^2} - \sqrt {0,2} .\sqrt {180{a^4}} \\
= \left( {9 - 6a + {a^2}} \right) - \sqrt {0,2.180.{a^4}} \\
= {a^2} - 6a + 9 - \sqrt {36{a^4}} \\
= {a^2} - 6a + 9 - 6{a^2}\\
= - 5{a^2} - 6a + 9\\
h,\\
\dfrac{{\sqrt {16{a^4}{b^6}} }}{{\sqrt {128{a^6}{b^2}} }} = \sqrt {\dfrac{{16{a^4}{b^6}}}{{128{a^6}{b^2}}}} = \sqrt {\dfrac{{{b^4}}}{{8{a^2}}}} = \sqrt {{{\left( {\dfrac{{{b^2}}}{{2\sqrt 2 a}}} \right)}^2}} \\
= \left| {\dfrac{{{b^2}}}{{2\sqrt 2 a}}} \right| = - \dfrac{{{b^2}}}{{2\sqrt 2 a}}\,\,\,\,\,\,\,\,\,\,\left( {a < 0} \right)
\end{array}\)
