Đáp án:
$b)M=\dfrac{\sqrt{x}-2}{\sqrt{x}-3} \\c)x= \dfrac{1}{4}$
Giải thích các bước giải:
$b)A=\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}\\ =\dfrac{x+3}{(\sqrt{x}+3)(\sqrt{x}-3)}+\dfrac{2}{\sqrt{x}+3}\\ =\dfrac{x+3}{(\sqrt{x}+3)(\sqrt{x}-3)}+\dfrac{2(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}\\ =\dfrac{x+3+2(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}\\ =\dfrac{x+2\sqrt{x}-3}{(\sqrt{x}+3)(\sqrt{x}-3)}\\ =\dfrac{x-\sqrt{x}+3\sqrt{x}-3}{(\sqrt{x}+3)(\sqrt{x}-3)}\\ =\dfrac{\sqrt{x}(\sqrt{x}-1)+3(\sqrt{x}-1)}{(\sqrt{x}+3)(\sqrt{x}-3)}\\ =\dfrac{(\sqrt{x}+3)(\sqrt{x}-1)}{(\sqrt{x}+3)(\sqrt{x}-3)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\\ M=A-B=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\\ c)M=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ \Leftrightarrow \dfrac{\sqrt{x}-2}{\sqrt{x}-3}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ \Leftrightarrow \dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=0\\ \Leftrightarrow \dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-3)(\sqrt{x}+2)}-\dfrac{(\sqrt{x}+1)(\sqrt{x}-3)}{(\sqrt{x}+2)(\sqrt{x}-3)}=0\\ \Leftrightarrow \dfrac{(\sqrt{x}-2)(\sqrt{x}+2)-(\sqrt{x}+1)(\sqrt{x}-3)}{(\sqrt{x}+2)(\sqrt{x}-3)}=0\\ \Leftrightarrow \dfrac{x-4-(x-2\sqrt{x}-3)}{(\sqrt{x}+2)(\sqrt{x}-3)}=0\\ \Leftrightarrow \dfrac{2\sqrt{x}-1}{(\sqrt{x}+2)(\sqrt{x}-3)}=0\\ \Leftrightarrow 2\sqrt{x}-1=0\\ \Leftrightarrow \sqrt{x}= \dfrac{1}{2} \Leftrightarrow x= \dfrac{1}{4}$
