Giải thích các bước giải:
Câu 1:
Ta có: $x=25\to\sqrt{x}=5$
$\to A=\dfrac{5+1}{5-1}=\dfrac64=\dfrac32$
Câu 2 :
Ta có:
$B=\dfrac{x+1}{x-\sqrt{x}} -\dfrac{2}{\sqrt{x}-1}$
$\to B=\dfrac{x+1}{\sqrt{x}(\sqrt{x}-1)} -\dfrac{2\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}$
$\to B=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}$
$\to B=\dfrac{(\sqrt{x}-1)^2}{\sqrt{x}(\sqrt{x}-1)}$
$\to B=\dfrac{\sqrt{x}-1}{\sqrt{x}}$
Câu 3:
Ta có:
$P=A\cdot B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot \dfrac{\sqrt{x}-1}{\sqrt{x}}$
$\to P=\dfrac{\sqrt{x}+1}{\sqrt{x}}$
$\to P=1+\dfrac{1}{\sqrt{x}}$
Mà $x>0\to \sqrt{x}>0\to \dfrac{1}{\sqrt{x}}>0\to 1+\dfrac{1}{\sqrt{x}}>1$
$\to P>1$
