Giải thích các bước giải:
Ta có :
$P=A:B=(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}):\dfrac{\sqrt{x}}{x+\sqrt{x}}$
$\to P=(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}).\dfrac{x+\sqrt{x}}{\sqrt{x}}$
$\to P=1+\dfrac{1}{\sqrt{x}}+\sqrt{x}$
$\to P\sqrt{x}=\sqrt{x}+1+x$
$\to \sqrt{x}+1+x+(2\sqrt{5}-1)\sqrt{x}=3x-2\sqrt{x-4}+3$
$\to 2\sqrt{5}.\sqrt{x}=2x-2\sqrt{x-4}+2$
$\to \sqrt{5}.\sqrt{x}=x-\sqrt{x-4}+1$
$\to 5x=(x-\sqrt{x-4}+1)^2$
Đặt $\sqrt{x-4}=t$
$\to 5(t^2+4)=(t^2+4-t+1)^2$
$\to t^4-2t^3+6t^2-10t+5=0$
$\to (t-1)^2(t^2+5)=0$
$\to t=1\to \sqrt{x-4}=1\to x=5$
