Giải thích các bước giải:
$a)P=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\\ Q=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\\ M=P.Q\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}+3}.\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\\ =\dfrac{\sqrt{x}+3-5}{\sqrt{x}+3}\\ =1-\dfrac{5}{\sqrt{x}+3}<1 \left(\text{Do }\dfrac{5}{\sqrt{x}+3}>0 \ \forall \ x \ge 0\right)$
Để $\sqrt{M}$ có nghĩa $\Rightarrow M \ge 0$
$0\le M<1 \Rightarrow M<\sqrt{M}\\ f)C=P.\dfrac{x+4\sqrt{x}+3}{\sqrt{x}}(x>0)\\=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}.\dfrac{x+\sqrt{x}+3\sqrt{x}+3}{\sqrt{x}}\\= \dfrac{\sqrt{x}+1}{\sqrt{x}+3}.\dfrac{\sqrt{x}(\sqrt{x}+1)+3(\sqrt{x}+1)}{\sqrt{x}}\\= \dfrac{\sqrt{x}+1}{\sqrt{x}+3}.\dfrac{(\sqrt{x}+3)(\sqrt{x}+1)}{\sqrt{x}}\\=\dfrac{(\sqrt{x}+1)^2}{\sqrt{x}}\\=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}\\=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}\\=\sqrt{x}+\dfrac{1}{\sqrt{x}}+2\\\ge 2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}+2(Cauchy)\\=4$
Dấu "=" xảy ra $\Leftrightarrow \sqrt{x}=\dfrac{1}{\sqrt{x}} \Leftrightarrow x=1$
