Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 4:\\ a.ĐK:\ a >0,\ a\neq 1\\ P=\ \frac{\sqrt{a} -1}{\sqrt{a} +1}\\ b.\ a=\{0;4;9\}\\ Bài\ 5:\\ a.\ ĐK\ x >0,\ x\neq 1\\ A=\frac{2-\sqrt{x}}{\sqrt{x} -1}\\ b.\ A=1\\ c.\ 1 >x >\frac{1}{4} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 4:\\ a.\ P=\left(\frac{a+2\sqrt{a}}{\sqrt{a} +2} -1\right) :\left(\frac{a-\sqrt{a}}{\sqrt{a} -1} +1\right)\\ ĐK:\ a >0,\ a\neq 1\\ P=\left(\sqrt{a} -1\right) :\left(\sqrt{a} +1\right) =\frac{\sqrt{a} -1}{\sqrt{a} +1}\\ b.\ P=1-\frac{2}{\sqrt{a} +1}\\ Để\ P\in \mathbb{Z} \Rightarrow \left(\sqrt{a} +1\right) =\{\pm 1;\pm 2\}\\ mà\ a\in \mathbb{Z} \Rightarrow a=\{4;0;1;9\} \ kết\ hợp\ ĐK\\ \Rightarrow a=\{0;4;9\}\\ Bài\ 5:\\ a.\ A=\frac{x\sqrt{x} +1}{x-1} -\frac{x-1}{\sqrt{x} +1}\\ ĐK\ x >0,\ x\neq 1\\ A=\frac{x\sqrt{x} +1-( x-1)\left(\sqrt{x} +1\right)}{\left(\sqrt{x} +1\right)\left(\sqrt{x} -1\right)}\\ A=\frac{x\sqrt{x} +1-x\sqrt{x} -x+\sqrt{x} +1}{\left(\sqrt{x} +1\right)\left(\sqrt{x} -1\right)}\\ A=\frac{-x+\sqrt{x} +2}{\left(\sqrt{x} +1\right)\left(\sqrt{x} -1\right)} =\frac{\left(\sqrt{x} +1\right)\left( 2-\sqrt{x}\right)}{\left(\sqrt{x} +1\right)\left(\sqrt{x} -1\right)}\\ A=\frac{2-\sqrt{x}}{\sqrt{x} -1}\\ b.\ x=\frac{9}{4} \Rightarrow A=1\\ c.\ A >1\\ \Leftrightarrow A-1=\frac{2-\sqrt{x}}{\sqrt{x} -1} -1=\frac{2-\sqrt{x} -\sqrt{x} +1}{\sqrt{x} -1}\\ =\frac{1-2\sqrt{x}}{\sqrt{x} -1} >0\\ TH\ 1:\ 1-2\sqrt{x} >0\ và\ \sqrt{x} -1 >0\\ \Leftrightarrow \sqrt{x} < \frac{1}{2} \ và\ \sqrt{x} >1\ ( vô\ lí)\\ TH\ 2:\ 1-2\sqrt{x} < 0\ và\ \sqrt{x} -1< 0\\ \Leftrightarrow 1 >\sqrt{x} >\frac{1}{2} \ \\ \Leftrightarrow 1 >x >\frac{1}{4} \ ( TM)\\ \end{array}$
