$\begin{array}{l}3)\,\,Q = \left(\dfrac{1}{\sqrt a + 1} - \dfrac{1}{a + \sqrt a}\right):\dfrac{\sqrt a - 1}{a + 2\sqrt a + 1}\\ a)\,\,ĐKXĐ:\,a >0\\ Q = \left(\dfrac{1}{\sqrt a + 1} - \dfrac{1}{\sqrt a(\sqrt a + 1)}\right)\cdot\dfrac{(\sqrt a + 1)^2}{\sqrt a - 1}\\ \to Q = \dfrac{\sqrt a - 1}{\sqrt a(\sqrt a + 1)}\cdot\dfrac{(\sqrt a + 1)^2}{\sqrt a - 1}\\ \to Q = \dfrac{\sqrt a + 1}{\sqrt a}\\ b)\,\,Xét\,\,Q - 1\\ = \dfrac{\sqrt a + 1}{\sqrt a} - 1\\ = \dfrac{\sqrt a + 1 - \sqrt a}{\sqrt a}\\ = \dfrac{1}{\sqrt a}\\ Do\,\,a >0\\ nên\,\,\sqrt a > 0\\ \to \dfrac{1}{\sqrt a} > 0\\ \to Q - 1 > 0\\ \to Q > 1\end{array}$
$4a)$ Ta có:
$\tan\widehat{B} = \dfrac{AH}{HB}$
$\to AH = HB.\tan\widehat{B} = 3.\tan60^o = 3\sqrt3 \, cm$
b) Ta có:
$\cos\widehat{B} = \dfrac{HB}{AB}$
$\to AB = \dfrac{HB}{\cos\widehat{B}} = \dfrac{3}{\cos60^o} = 6\, cm$
c) Áp dụng định lý Pytago vào $ΔHAC$ vuông tại $H$ ta được:
$AC^2 = AH^2 + HC^2$
$\Rightarrow AC = \sqrt{AH^2+HC^2} = \sqrt{(3\sqrt3)^2 + 4^2} = \sqrt{43}\, cm$
d) Ta có:
$\tan\widehat{C} = \dfrac{AH}{HC} = \dfrac{3\sqrt3}{4}$
$\to \widehat{C} = \arctan\dfrac{3\sqrt3}{4} \approx 52,41^o$
