Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\\ a.\ Q=-1\\ b.\ P=\frac{2\sqrt{x} -2}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}\\ c.\ GTNN\ cuả\ \frac{P}{Q} =1\\ Bài\ 2:\\ a.\ P=\frac{x}{\sqrt{x} -1}\\ b.\ x=4+2\sqrt{3} \ \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\\ a.\ Q=\frac{3-1}{1-2} +1=-1\\ b.\ P=\frac{\sqrt{x}\left(\sqrt{x} +2\right) -\left(\sqrt{x} +1\right)\left(\sqrt{x} -2\right) -\sqrt{x} -4}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}\\ =\frac{x+2\sqrt{x} -x+\sqrt{x} +2-\sqrt{x} -4}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}\\ =\frac{2\sqrt{x} -2}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}\\ c.\ A=P:Q=\frac{2\sqrt{x} -2}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)} .\frac{\sqrt{x} -2}{3-\sqrt{x} +\sqrt{x} -2}\\ =\frac{2\sqrt{x} -2}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)} .\frac{\sqrt{x} -2}{1} =\frac{2\sqrt{x} -2}{\sqrt{x} +2}\\ Ta\ có\ A+1=\frac{2\sqrt{x} -2}{\sqrt{x} +2} +1=\frac{2\sqrt{x} -2+\sqrt{x} +2}{\sqrt{x} +2}\\ =\frac{3\sqrt{x}}{\sqrt{x} +2} \geqslant 0\\ \Rightarrow A\geqslant 1\\ Vâỵ\ GTNN\ cuả\ \frac{P}{Q} =1\\ Bài\ 2:\\ a.\ P=\frac{\sqrt{x}\left(\sqrt{x} +1\right) +\sqrt{x}}{\left(\sqrt{x} +1\right)\left(\sqrt{x} -1\right)} .\frac{x\left(\sqrt{x} +1\right)}{2\left(\sqrt{x} +1\right) -2+x}\\ P=\frac{x+2\sqrt{x}}{\left(\sqrt{x} +1\right)\left(\sqrt{x} -1\right)} .\frac{x\left(\sqrt{x} +1\right)}{2\sqrt{x} +2-2+x}\\ P=\frac{x+2\sqrt{x}}{\left(\sqrt{x} +1\right)\left(\sqrt{x} -1\right)} .\frac{x\left(\sqrt{x} +1\right)}{2\sqrt{x} +x}\\ P=\frac{x}{\sqrt{x} -1}\\ b.\ P=2\\ \Rightarrow \frac{x}{\sqrt{x} -1} =2\Leftrightarrow x-2\sqrt{x} -2=0\\ \Leftrightarrow \sqrt{x} =1+\sqrt{3}\\ \Leftrightarrow x=4+2\sqrt{3} \ ( TM) \end{array}$
