Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Câu\ \ 1:\\ a.\\ B=\frac{2\sqrt{x} +3}{\left(\sqrt{x} -2\right)\left( 2\sqrt{x} +1\right)}\\ b.\ x=\{\emptyset \}\\ Câu\ 2:\\ 1.\ x=9\ \\ 2.\ A.B=1\\ 3.\ x >1\ \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Câu\ \ 1:\\ a.\ B=\frac{2\left( 2\sqrt{x} +1\right) +3\left(\sqrt{x} -2\right) -5\sqrt{x} +7}{\left(\sqrt{x} -2\right)\left( 2\sqrt{x} +1\right)}\\ B=\frac{2\sqrt{x} +3}{\left(\sqrt{x} -2\right)\left( 2\sqrt{x} +1\right)}\\ b.\ C=\frac{2\sqrt{x} +3}{\left(\sqrt{x} -2\right)\left( 2\sqrt{x} +1\right)} :\frac{5\sqrt{x}\left(\sqrt{x} -2\right)}{2\sqrt{x} +3}\\ C=\frac{5\sqrt{x} -10}{2\sqrt{x} +1} =\frac{5\sqrt{x}}{2\sqrt{x} +1} -\frac{10}{2\sqrt{x} +1}\\ Để\ C\in \mathbb{Z} \Rightarrow \frac{5\sqrt{x}}{2\sqrt{x} +1} \in \mathbb{Z} \ và\ \frac{10}{2\sqrt{x} +1} \in \mathbb{Z}\\ +) \ \frac{10}{2\sqrt{x} +1} \in \mathbb{Z}\\ \Rightarrow 2\sqrt{x} +1=\{1;2;5;10\}\\ \Rightarrow x=\left\{0;\frac{1}{4} ;4;\frac{81}{4}\right\} ;\ thay\ vào\ \frac{5\sqrt{x}}{2\sqrt{x} +1} \in \mathbb{Z}\\ \Rightarrow x=\{0;4\} \ kết\ hợp\ với\ ĐKXĐ\\ \Rightarrow x=\{\emptyset \}\\ Câu\ 2:\\ 1.\ A=2\\ \Rightarrow \sqrt{x} +1=2\sqrt{x} -1\\ \Leftrightarrow \sqrt{x} =3\\ \Leftrightarrow x=9\ ( TM)\\ 2.\ A.B=\frac{\sqrt{x} +1}{\sqrt{x} -1} .\frac{\sqrt{x}\left(\sqrt{x} -1\right) -\sqrt{x} +1}{\left(\sqrt{x} -1\right)\left(\sqrt{x} +1\right)}\\ =\frac{\sqrt{x} +1}{\sqrt{x} -1} .\frac{x-\sqrt{x} -\sqrt{x} +1}{\left(\sqrt{x} -1\right)\left(\sqrt{x} +1\right)}\\ =\frac{\sqrt{x} +1}{\sqrt{x} -1} .\frac{x -2\sqrt{x} +1}{\left(\sqrt{x} -1\right)\left(\sqrt{x} +1\right)}\\ =\frac{\sqrt{x} +1}{\sqrt{x} -1} .\frac{\left(\sqrt{x} - 1\right)^{2}}{\left(\sqrt{x} -1\right)\left(\sqrt{x} +1\right)}\\ =1\\ 3.\ A\geqslant B\\ \Leftrightarrow A-B\geqslant 0\\ \Leftrightarrow \frac{\sqrt{x} +1}{\sqrt{x} -1} -\frac{\left(\sqrt{x} - 1\right)^{2}}{\left(\sqrt{x} -1\right)\left(\sqrt{x} +1\right)} \geqslant 0\\ \Leftrightarrow \frac{\sqrt{x} +1}{\sqrt{x} -1} -\frac{\sqrt{x} - 1}{\sqrt{x} +1} \geqslant 0\\ \Leftrightarrow \frac{x+2\sqrt{x} +1-x+2\sqrt{x} -1}{x-1} \geqslant 0\\ \Leftrightarrow \frac{4\sqrt{x}}{x-1} \geqslant 0\\ \Leftrightarrow x-1 >0\\ \Leftrightarrow x >1\ ( TM) \end{array}$
