Đáp án:
a/ $B=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}$
b/ $M_{MIN}=2\sqrt{6}$ khi $x=6$
Giải thích các bước giải:
a/ $B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}$ $(1)$
$\text{ĐKXĐ: $x > 0$}$
$(1)=\dfrac{(\sqrt{x}-1)(\sqrt{x}+1)+2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}$
$=\dfrac{x-1+2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}$
$=\dfrac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}+1)}$
$=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}$
b/ Ta có: $\dfrac{A}{B}=\dfrac{2+\sqrt{x}}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}$
$=\dfrac{2+\sqrt{x}}{\sqrt{x}}.\dfrac{\sqrt{x}+1}{\sqrt{x}+2}$
$=\dfrac{\sqrt{x}+1}{\sqrt{x}}$
$⇒ M=\dfrac{A}{B}+\dfrac{x+5-\sqrt{x}}{\sqrt{x}}$ $(2)$
$\text{ĐKXĐ: $x > 0$}$
$(2)=\dfrac{\sqrt{x}+1}{\sqrt{x}}+\dfrac{x+5-\sqrt{x}}{\sqrt{x}}$
$=\dfrac{\sqrt{x}+1+x+5-\sqrt{x}}{\sqrt{x}}$
$=\dfrac{x+6}{\sqrt{x}}$
$=\sqrt{x}+\dfrac{6}{\sqrt{x}}$
$\text{Áp dụng bất đẳng thức Cô-si cho 2 số dương ta có:}$
$\sqrt{x}+\dfrac{6}{\sqrt{x}} \geq 2\sqrt{\sqrt{x}.\dfrac{6}{\sqrt{x}}}=2\sqrt{6}$
$\text{Dấu "=" xảy ra khi $\sqrt{x}=\dfrac{6}{\sqrt{x}} ⇔ x=6$}$
$\text{Vậy $M_{MIN}=2\sqrt{6}$ khi $x=6$}$
