a) Ta có:
\( \sqrt{0,36a^{2}}\ = \sqrt{0,36}.\sqrt{a^{2}}\)
\(=\sqrt{0,6^2}.\sqrt{a^2}\)
\(= 0,6.│a│\)
\(= 0,6. (-a)=-0,6a\)
(Vì \(a < 0\) nên \(│a│= -a)\).
b)
Vì \( a^{2}\) ≥ 0 nên \(\left| a^2 \right|= a^{2}\).
Vì \(a \ge 3\) hay \(3 \le a \) nên \(3 - a ≤ 0\).
\( \Rightarrow│3 - a│= -(3-a)=-3+a=a - 3\).
Ta có: \( \sqrt{a^{4}.(3 - a)^{2}}= \sqrt{a^{4}}\).\( \sqrt{(3 - a)^{2}}\)
\(=\sqrt{(a^2)^2}.\sqrt{(3-a)^2}\)
\(= \left| a^{2}\right|.\left| 3 - a \right|\).
\(= a^2.(a-3)=a^3-3a^2\).
c)
Vì \(a > 1\) hay \(1<a\) nên \(1 - a < 0\).
\( \Rightarrow \left| 1 - a\right| =-(1-a)=-1+a= a -1\).
Ta có: \( \sqrt{27.48(1 - a)^{2}} = \sqrt{27.(3.16).(1 - a)^{2}}\)
\(=\sqrt{(27.3).16.(1-a)^2}\)
\(= \sqrt{81.16.(1 - a)^{2}}\)
\(=\sqrt {81} .\sqrt {16} .\sqrt {{{(1 - a)}^2}} \)
\(=\sqrt{9^2}.\sqrt{4^2}.\sqrt{(1-a)^2}\)
\(= 9.4. \left| {1 - a} \right| = 36.\left| {1 - a} \right|\)
\(= 36.(a-1)=36a-36\).
d)
Vì \(a^2 \ge 0\), với mọi \(a\) nên \( \left|a^2 \right| = a^2\).
Vì \(a > b\) nên \(a -b > 0\). Do đó \(\left|a - b\right|= a - b\).
Ta có: \( \dfrac{1}{a - b}\) . \( \sqrt{a^{4}.(a - b)^{2}}\)
\(= \dfrac{1}{a - b}\) . \( \sqrt{a^{4}}.\sqrt{(a - b)^{2}}\)
\(= \dfrac{1}{a - b} . \left( {\left| {{a^2}} \right|.\left| {a - b} \right|} \right)\)
\(=\dfrac{1}{a - b} . [ a^{2}(a - b)] \)
\(=a^2\)
