a) Ta có:
\(\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{2a}{3}.\dfrac{3a}{8}}=\sqrt{\dfrac{2a.3a}{3.8}}\) \(=\sqrt{\dfrac{a^2}{4}}=\sqrt{\dfrac{a^2}{2^2}}\)
\(=\sqrt{\left(\dfrac{a}{2}\right)^2}=\left| \dfrac{a}{2}\right|\) \(= \dfrac{a}{2}\).
(Vì \(a \ge 0\) nên \(\dfrac{a}{2} \ge 0 \) \( \Rightarrow \left| \dfrac{a}{2} \right| = \dfrac{a}{2}\)).
b) Ta có:
\(\sqrt{13a}.\sqrt{\dfrac{52}{a}}=\sqrt{13a.\dfrac{52}{a}}=\sqrt{\dfrac{13a.52}{a}}\)
\(=\sqrt{\dfrac{13a.(13.4)}{a}}=\sqrt{\dfrac{(13.13).4.a}{a}}\)
\(=\sqrt{13^2.4}=\sqrt{13^2}.\sqrt{4}\)
\(=\sqrt{13^2}.\sqrt{2^2}=13.2\)
\(=26\) (vì \(a>0\))
c)
Do \(a\geq 0\) nên bài toán luôn được xác định có nghĩa.
Ta có: \(\sqrt{5a}.\sqrt{45a}- 3a=\sqrt{5a.45a}-3a\)
\(=\sqrt{(5.a).(5.9.a)}-3a\)
\(=\sqrt{(5.5).9.(a.a)}-3a\)
\(=\sqrt{5^2.3^2.a^2}-3a\)
\(=\sqrt{5^2}.\sqrt{3^2}.\sqrt{a^2}-3a\)
\(=5.3.\left|a\right|-3a=15 \left|a \right| -3a.\)
\(=15a - 3a = (15-3)a =12a.\)
Vì \(a \ge 0\) nên \(\left| a \right| = a.\)
d) Ta có:
\((3 - a)^{2}- \sqrt{0,2}.\sqrt{180a^{2}}=\sqrt{0,2.180a^2}\)
\(= (3-a)^2-\sqrt{0,2.(10.18).a^2}\)
\(=(3-a)^2-\sqrt{(0,2.10).18.a^2}\)
\(=(3-a)^3-\sqrt{2.18.a^2}\)
\(=(3-a)^2-\sqrt{36a^2}\)
\(=(3-a)^2-\sqrt{36}.\sqrt{a^2}\)
\(=(3-a)^2-\sqrt{6^2}.\sqrt{a^2}\)
\(=(3-a)^2-6.\left|a\right|\).
+) \(TH1\): Nếu \(a\geq 0\Rightarrow |a|=a\).
Do đó: \((3 - a)^{2}- 6\left|a\right|=(3-a)^2-6a\)
\(=(3^2-2.3.a+a^2)-6a\)
\(=(9-6a+a^2)-6a\)
\(=9-6a+a^2-6a\)
\(=a^2+(-6a-6a)+9\)
\(=a^2+(-12a)+9\)
\(=a^2-12a+9\).
+) \(TH2\): Nếu \(a<0\Rightarrow |a|=-a\).
Do đó: \((3 - a)^{2}- 6\left|a\right| =(3-a)^2-6.(-a)\)
\(=(3^2-2.3.a+a^2)-(-6a)\)
\(=(9-6a+a^2)+6a\)
\(=9-6a+a^2+6a\)
\(=a^2+(-6a+6a)+9\)
\(=a^2+9\).
Vậy \((3 - a)^{2}- \sqrt{0,2}.\sqrt{180a^{2}}=a^2-12a+9\), nếu \(a \ge 0\).
\((3 - a)^{2}- \sqrt{0,2}.\sqrt{180a^{2}}=a^2+9\), nếu \(a <0\).
