Giải thích các bước giải:
b.Ta có $x=9\to \sqrt{x}=3$
$\to B= \dfrac{3-2}3=\dfrac13$
b.Ta có:
$A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+3}+\dfrac{4}{x+3\sqrt{x}}-1$
$\to A=\dfrac{\sqrt{x}(2\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}+3)}+\dfrac{4}{\sqrt{x}(\sqrt{x}+3)}-1$
$\to A=\dfrac{\sqrt{x}(2\sqrt{x}-1)+4}{\sqrt{x}(\sqrt{x}+3)}-1$
$\to A=\dfrac{2x-\sqrt{x}+4}{\sqrt{x}(\sqrt{x}+3)}-1$
$\to A=\dfrac{2x-\sqrt{x}+4-\sqrt{x}(\sqrt{x}+3)}{\sqrt{x}(\sqrt{x}+3)}$
$\to A=\dfrac{2x-\sqrt{x}+4-(x+3\sqrt{x})}{\sqrt{x}(\sqrt{x}+3)}$
$\to A=\dfrac{2x-\sqrt{x}+4-x-3\sqrt{x}}{\sqrt{x}(\sqrt{x}+3)}$
$\to A=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}(\sqrt{x}+3)}$
$\to A=\dfrac{(\sqrt{x}-2)^2}{\sqrt{x}(\sqrt{x}+3)}$
c.Ta có:
$A:B=m$
$\to \dfrac{(\sqrt{x}-2)^2}{\sqrt{x}(\sqrt{x}+3)}:\dfrac{\sqrt{x}-2}{\sqrt{x}}=m$
$\to \dfrac{(\sqrt{x}-2)^2}{\sqrt{x}(\sqrt{x}+3)}\cdot \dfrac{\sqrt{x}}{\sqrt{x}-2}=m$
$\to \dfrac{\sqrt{x}-2}{\sqrt{x}+3}=m$
$\to \sqrt{x}-2=m(\sqrt{x}+3)$
$\to \sqrt{x}-2=m\sqrt{x}+3m$
$\to (m-1)\sqrt{x}=-(3m+2)$
Nếu $m=1\to (1-1)\cdot \sqrt{x}=-(3\cdot 1+2)\to 0=-5$ vô lý
$\to m=1$ loại
Nếu $m\ne 1$
$\to \sqrt{x}=-\dfrac{3m+2}{m-1}(*)$
Để phương trình có nghiệm duy nhất
$\to (*)$ có nghiệm duy nhất
Mà $\sqrt{x}>0$ do $x>0$
$\to -\dfrac{3m+2}{m-1}>0$
$\to \dfrac{3m+2}{m-1}<0$
$\to -\dfrac23<m<1$
