a) Ta có:
\(\sqrt{18(\sqrt{2}-\sqrt{3})^{2}}=\sqrt {18}.\sqrt{(\sqrt 2 - \sqrt 3)^2}\)
\(=\sqrt{9.2}.|\sqrt{2}-\sqrt{3}|=\sqrt{3^2.2}.|\sqrt{2}-\sqrt{3}|\)
\(=3\sqrt{2}.|\sqrt{2}-\sqrt{3}|=3\sqrt{2}(\sqrt{3}-\sqrt{2})\)
\(=3\sqrt {2.3}- 3(\sqrt 2)^2\)
\(=3\sqrt 6 -3.2=3\sqrt{6}-6\).
(Vì \( 2 < 3 \Leftrightarrow \sqrt 2 < \sqrt 3 \Leftrightarrow \sqrt 2 -\sqrt 3 <0\)
Do đó: \( |\sqrt 2 -\sqrt 3|=-(\sqrt 2 -\sqrt 3)=-\sqrt 2 +\sqrt 3\)\(=\sqrt 3-\sqrt2\)).
b) Ta có:
\(ab\sqrt{1+\dfrac{1}{a^{2}b^{2}}}=ab\sqrt{\dfrac{a^2b^2}{a^2b^2}+\dfrac{1}{a^2b^2}}=ab\sqrt{\dfrac{a^2b^2+1}{a^2b^2}}\)
\(=ab\dfrac{\sqrt{a^2b^2+1}}{\sqrt{a^2b^2}}=ab\dfrac{\sqrt{a^2b^2+1}}{\sqrt{(ab)^2}}\)
\(=ab\dfrac{\sqrt{a^2b^2+1}}{|ab|}\)
Nếu \(ab \ge 0\) thì \(|ab|=ab\)
\( \Rightarrow ab\dfrac{\sqrt{a^2b^2+1}}{|ab|}=ab\dfrac{\sqrt{a^2b^2+1}}{ab}=\sqrt{a^2b^2+1}\).
Nếu \(ab < 0\) thì \(|ab|=-ab \)
\(\Rightarrow ab\dfrac{\sqrt{a^2b^2+1}}{|ab|}=ab\dfrac{\sqrt{a^2b^2+1}}{-ab}=-\sqrt{a^2b^2+1}\).
c) Ta có:
\(\sqrt{\dfrac{a}{b^{3}}+\dfrac{a}{b^{4}}}=\sqrt{\dfrac{a.b}{b^{3}.b}+\dfrac{a}{b^{4}}}=\sqrt{\dfrac{ab}{b^4}+\dfrac{a}{b^4}}\)
\(=\sqrt{\dfrac{ab+a}{b^4}}=\dfrac{\sqrt{ab+a}}{\sqrt{(b^2)^2}}=\dfrac{\sqrt{ab+a}}{|b^2|}=\dfrac{\sqrt{ab+a}}{b^2}\).
(Vì \(b^2 > 0\) với mọi \( b \ne 0\) nên \( |b^2|=b^2\)).
d) Ta có:
\(\dfrac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{(\sqrt a)^2+\sqrt{a}.\sqrt b}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt a (\sqrt a+\sqrt b)}{\sqrt{a}+\sqrt{b}}\)
\(=\sqrt a\).
