$\begin{array}{l}1a) \, Q = \left(\dfrac{\sqrt x - 2}{x - 1}-\dfrac{\sqrt x + 2}{x + 2\sqrt x + 1}\right)\cdot\left(\dfrac{1 - x}{\sqrt 2}\right)^2\\ = \left[\dfrac{(\sqrt x - 2)(x - 1)^2}{x - 1} - \dfrac{(\sqrt x + 2)(x - 1)^2}{(\sqrt x + 1)^2}\right]\cdot \dfrac{1}{2}\\ = \left[(\sqrt x - 2)(x - 1) - (\sqrt x + 2)(\sqrt x - 1)^2\right]\cdot\dfrac{1}{2}\\ =\left[x\sqrt x - \sqrt x - 2x + 2- (\sqrt x + 2)(x - 2\sqrt x + 1) \right]\cdot \dfrac{1}{2}\\ = \dfrac{1}{2}\cdot(x\sqrt x - \sqrt x - 2x + 2- x\sqrt x + 2x - \sqrt x - 2x + 4\sqrt x - 2)\\ = \dfrac{1}{2}\cdot(2x + 2\sqrt x)\\ = x + \sqrt x\\ b) \,\text{Q không có giá trị lớn nhất}\\ Ta\,\,có:\\ x \geq 0\\ \sqrt x \geq 0\\ \Rightarrow x + \sqrt x \geq 0\\ \Rightarrow Q \geq 0\\ minQ = 0 \Leftrightarrow \sqrt x = 0 \Leftrightarrow x = 0\end{array}$
$2)$ Từ $C$ dựng $\widehat{DCE}$ sao cho $\widehat{DCE} = \widehat{BAD} \, (E \in AD)$
Xét $ΔBAD$ và $ΔECD$ có:
$\widehat{BAD} = \widehat{DCE}$ (cách dựng)
$\widehat{BDA} = \widehat{EDC}$ (đối đỉnh)
Do đó $ΔBAD \sim ΔECD \, (g.g)$
$\Rightarrow \dfrac{BD}{DE} = \dfrac{AD}{DC}$
$\Rightarrow BD.DC = AD.DE$ $(1)$
Xét $ΔBAD$ và $ΔEAC$ có:
$\widehat{BAD} = \widehat{EAC}\, (gt)$
$\widehat{ABD} = \widehat{AEC} \, (ΔBAD \sim ΔECD)$
Do đó $ΔBAD\sim ΔEAC \, (g.g)$
$\Rightarrow \dfrac{AD}{AC} = \dfrac{AB}{AE}$
$\Rightarrow AB.AC = AD.AE$ $(2)$
Lấy $(2) - (1)$ ta được:
$AB.AC - BD.DC = AD.AE - AD.DE = AD(AE - DE) = AD.AD = AD^2$
Vậy $AD^2 = AB.AC - BD.DC$
