Đáp án:
$1)\\ a)B=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\\ b)0<x<1\\ c) x \in \varnothing\\ d) B>1\\ e) max_B=\dfrac{5}{2}\\ 2)\\ a)A=\dfrac{\sqrt{x}}{\sqrt{x}+1} \ \ \ \ ĐKXĐ: \left\{\begin{array}{l} x \ge 0\\x\ne 1\end{array} \right.\\ B=\dfrac{\sqrt{x}-2}{\sqrt{x}} \ \ \ \ ĐKXĐ: x>0\\ b)A>B$
Giải thích các bước giải:
$1)B=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{7}{x-4}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)(x \ge 0 ; x \ne 4)\\ a)B=\left(\dfrac{\sqrt{x}-2}{(\sqrt{x}+2)(\sqrt{x}-2)}+\dfrac{7}{(\sqrt{x}+2)(\sqrt{x}-2)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-2}\right)\\ =\dfrac{\sqrt{x}-2+7}{(\sqrt{x}+2)(\sqrt{x}-2)}:\dfrac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\\ =\dfrac{\sqrt{x}+5}{(\sqrt{x}+2)(\sqrt{x}-2)}.(\sqrt{x}-2)\\ =\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\\ b)B=\dfrac{\sqrt{x}+5}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+2+3}{\sqrt{x}+2}\\ =1+\dfrac{3}{\sqrt{x}+2}\\ B>2\Leftrightarrow 1+\dfrac{3}{\sqrt{x}+2}>2\\ \Leftrightarrow \dfrac{3}{\sqrt{x}+2}>1\\ \Leftrightarrow \dfrac{3}{\sqrt{x}+2}-1>0\\ \Leftrightarrow \dfrac{3-\sqrt{x}-2}{\sqrt{x}+2}>0\\ \Leftrightarrow \dfrac{1-\sqrt{x}}{\sqrt{x}+2}>0\\ \Leftrightarrow 1-\sqrt{x}>0\\ \Leftrightarrow x<1$
Kết hợp điều kiện $\Rightarrow 0<x<1$
$c)B=\dfrac{2}{3}\\ \Leftrightarrow \dfrac{\sqrt{x}+5}{\sqrt{x}+2}=\dfrac{2}{3}\\ \Leftrightarrow \dfrac{\sqrt{x}+5}{\sqrt{x}+2}-\dfrac{2}{3}=0\\ \Leftrightarrow \dfrac{3\sqrt{x}+15-2(\sqrt{x}+2)}{\sqrt{x}+2}=0\\ \Leftrightarrow \dfrac{\sqrt{x}+11}{\sqrt{x}+2}=0\\ \Leftrightarrow x \in \varnothing\\ d)B=1+\underbrace{\dfrac{3}{\sqrt{x}+2}}_{>0 \ \forall \ x \ge 0 ; x \ne 4} >1 \ \forall \ x \ge 0 ; x \ne 4\\ e)B=1+\dfrac{3}{\sqrt{x}+2}\\ B \ max \Leftrightarrow \dfrac{3}{\sqrt{x}+2} \ max$
Do $\sqrt{x}+2 >0 \ \forall \ x \ge 0 ; x \ne 4$ nên $\dfrac{3}{\sqrt{x}+2} \ max \Leftrightarrow \sqrt{x}+2 \ min $
$\sqrt{x}+2 \ge 2 \ \forall \ x \ge 0 ; x \ne 4 \\ \Rightarrow max_B=1+\dfrac{3}{2}=\dfrac{5}{2}$
Dấu "=" xảy ra $\Leftrightarrow x=0.$
$2)a)A=\dfrac{x-\sqrt{x}}{x-1} \ \ \ \ ĐKXĐ: \left\{\begin{array}{l} x \ge 0\\ x \ne 1\end{array} \right.\\ B=\dfrac{x-4}{x+2\sqrt{x}}\\ =\dfrac{x-4}{\sqrt{x}(\sqrt{x}+2)} \ \ \ \ ĐKXĐ: \left\{\begin{array}{l} x \ge 0\\ \sqrt{x}(\sqrt{x}+2) \ne 0\end{array} \right. \Leftrightarrow x>0\\ A=\dfrac{x-\sqrt{x}}{x-1}\\ =\dfrac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\\ =\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ B=\dfrac{x-4}{\sqrt{x}(\sqrt{x}+2)}\\ =\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}+2)}\\ =\dfrac{\sqrt{x}-2}{\sqrt{x}}\\ b)A-B=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{\sqrt{x}}\\ =\dfrac{\sqrt{x}.\sqrt{x}-(\sqrt{x}-2)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}(\sqrt{x}+1)} >0 \ \forall \ x>0\\\Rightarrow A>B$.
