a) Ta có:
\(ab^{2}.\sqrt{\dfrac{3}{a^{2}b^{4}}}=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}\) \(=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2}.\sqrt{b^4}}\)
\(=ab^2.\dfrac{\sqrt{3}}{\sqrt{a^2}.\sqrt{(b^2)^2}}\) \(=ab^2.\dfrac{\sqrt{3}}{|a|.|b^2|}\)
\(=ab^2.\dfrac{\sqrt{3}}{-ab^2}=-\sqrt{3}\).
(Vì \(a < 0 \) nên \(|a|=-a\) và \(b \ne 0\) nên \(b^2 >0 \Rightarrow |b^2|=b^2) \).
b) Ta có:
\(\sqrt{\dfrac{27(a - 3)^{2}}{48}}=\sqrt{\dfrac{27}{48}.(a-3)^2}\) \(=\sqrt{\dfrac{27}{48}}.\sqrt{(a-3)^2}\)
\(=\sqrt{\dfrac{9.3}{16.3}}.\sqrt{(a-3)^2}\) \(=\sqrt{\dfrac{9}{16}}.\sqrt{(a-3)^2}\)
\(=\sqrt{\dfrac{3^2}{4^2}}.\sqrt{(a-3)^2}\) \(=\dfrac{\sqrt {3^2}}{\sqrt {4^2}}.\sqrt{(a-3)^2}\)
\(=\dfrac{3}{4}|a-3|=\dfrac{3}{4}(a-3)\).
( Vì \(a > 3\) nên \(a-3>0 \Rightarrow |a-3|=a-3) \)
c) Ta có:
\(\sqrt{\dfrac{9+12a+4a^{2}}{b^{2}}}=\sqrt{\dfrac{3^2+2.3.2a+2^2.a^2}{b^2}}\)
\(=\sqrt{\dfrac{3^2+2.3.2a+(2a)^2}{b^2}}=\sqrt{\dfrac{(3+2a)^2}{b^2}}\)
\(=\dfrac{\sqrt{(3+2a)^2}}{\sqrt{b^2}}=\dfrac{|3+2a|}{|b|}\)
Vì \(a \geq -1,5 \Rightarrow a+1,5>0\)
\(\Leftrightarrow 2(a+1,5)>0\)
\( \Leftrightarrow 2a+3>0\)
\( \Leftrightarrow 3+2a>0\)
\(\Rightarrow |3+2a|=3+2a\)
Vì \(b<0\Rightarrow |b|=-b\)
Do đó: \(\dfrac{|3+2a|}{|b|}=\dfrac{3+2a}{-b} =-\dfrac{3+2a}{b}\).
Vậy \(\sqrt{\dfrac{9+12a+4a^{2}}{b^{2}}}=-\dfrac{3+2a}{b}\).
d) Ta có:
\((a - b).\sqrt{\dfrac{ab}{(a - b)^{2}}}=(a-b).\dfrac{\sqrt{ab}}{\sqrt{(a-b)^2}}\)
\(=(a-b).\dfrac{\sqrt{ab}}{|a-b|}\)
\(=(a-b).\dfrac{\sqrt{ab}}{-(a-b)}=-\sqrt{ab}\).
(Vì \(a < b < 0\) nên \(a-b<0\Rightarrow |a-b|=-(a-b))\).
