Bài 4 :
a)$\frac{2\sqrt{x}-3}{\sqrt{x}-2}+\frac{\sqrt{x}}{x-2\sqrt{x}}+1$
$=\frac{2\sqrt{x}-3}{\sqrt{x}-2}+\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-2)}+1$
$=\frac{\sqrt{x}(2\sqrt{x}-3)+\sqrt{x}}{\sqrt{x}(\sqrt{x}-2)}+1$
$=\frac{2x-3\sqrt{x}+\sqrt{x}}{\sqrt{x}(\sqrt{x}-2)}+1$
$=\frac{2x-2\sqrt{x}}{\sqrt{x}(\sqrt{x}-2)}+1$
$=\frac{2\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-2)}+1$
$=\frac{2(\sqrt{x}-1)}{\sqrt{x}-2}+1$
$=\frac{2(\sqrt{x}-1)}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}-2}$
$=\frac{2\sqrt{x}-2+\sqrt{x}-2}{\sqrt{x}-2}$
$=\frac{3\sqrt{x}-4}{\sqrt{x}-2}$
b) Thay x = 1 vào A ta được :
$A=\frac{3\sqrt{1}-4}{\sqrt{1}-2}$
$=\frac{-1}{-1}$ $=1$
c) $\frac{3\sqrt{x}-4}{\sqrt{x}-2}$ nguyên
=> $3\sqrt{x}-4:\sqrt{x}-2$
<=>$2\sqrt{x}:2$
=> $x∈Ư(2)$ => $x=(-1,-2,1,2)$
=> $\sqrt{x}∈ ( 3 , 4 , 1, 0 )$
mà $x>0$
=> $x∈(1,9,16)$
