Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\\ 1.\ P=\frac{4\sqrt{x}}{\sqrt{x} +3}\\ 2.\ P=\frac{8}{5}\\ 3.\ x=9\ \\ Bài\ 5:\\ 1.\ P=\frac{x+\sqrt{x} +1}{\sqrt{x}}\\ 2.\ P=\frac{7}{2}\\ Bài\ 7:\\ 1.\ P=1-2\sqrt{x}\\ 2.\ x=1\ \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\\ 1.\ P=\frac{\left(\sqrt{x} +2\right)^{2} -\left(\sqrt{x} -2\right)^{2} +4x}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)} .\frac{x-4}{\left(\sqrt{x} +2\right)\left(\sqrt{x} +3\right)}\\ P=\frac{x+4\sqrt{x} +4-x+4\sqrt{x} -4+4x}{\left(\sqrt{x} +2\right)\left(\sqrt{x} +3\right)}\\ P=\frac{8\sqrt{x} +4x}{\left(\sqrt{x} +2\right)\left(\sqrt{x} +3\right)} =\frac{4\sqrt{x}\left(\sqrt{x} +2\right)}{\left(\sqrt{x} +2\right)\left(\sqrt{x} +3\right)} =\frac{4\sqrt{x}}{\sqrt{x} +3}\\ 2.\ x=\sqrt{\left(\sqrt{5} +2\right)^{2}} -\sqrt{\left(\sqrt{5} -2\right)^{2}} =\sqrt{5} +2-\sqrt{5} +2=4\\ \Rightarrow P=\frac{4\sqrt{4}}{\sqrt{4} +3} =\frac{8}{5}\\ 3.\\ P=2\Leftrightarrow \frac{4\sqrt{x}}{\sqrt{x} +3} =2\Leftrightarrow 4\sqrt{x} =2\sqrt{x} +6\Leftrightarrow x=9\ ( TM)\\ Bài\ 5:\\ 1.\ P=\frac{x+\sqrt{x} +1}{\left(\sqrt{x} +1\right)\sqrt{x}} .\frac{x+\sqrt{x}}{\sqrt{x}} =\frac{x+\sqrt{x} +1}{\sqrt{x}}\\ 2.\ x=4\Rightarrow P=\frac{4+\sqrt{4} +1}{\sqrt{4}} =\frac{4+2+1}{2} =\frac{7}{2}\\ Bài\ 7:\\ 1.\ P=\frac{\left(\sqrt{x} -2\right)\left( x+2\sqrt{x} +4\right)}{x+2\sqrt{x} +4} +3-3\sqrt{x}\\ P=\sqrt{x} -2+3-3\sqrt{x} =1-2\sqrt{x}\\ 2.\ Q=\frac{2P}{1-P} =\frac{2-4\sqrt{x}}{2\sqrt{x}} =-2+\frac{1}{\sqrt{x}}\\ Để\ Q\in \mathbb{Z} \Rightarrow \frac{1}{\sqrt{x}} \in \mathbb{Z}\\ \Rightarrow \sqrt{x} =\{1\}\\ \Rightarrow x=1\ ( TM)\\ \end{array}$
