Đáp án:a)\(\frac{5}{6}\)
b)\(\frac{\sqrt{x}-2}{\sqrt{x}-3}\)
c)\(x=\frac{49}{4}\)
Giải thích các bước giải:
đk: \(x\neq9, x>0\)
b)A=\(\frac{x+3}{x-9}+\frac{2}{\sqrt{x}+3}-\frac{1}{\sqrt{x}-3}\)
=\(\frac{x+3+2(\sqrt{x}-3)-\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}\)
=\(\frac{x+3+2\sqrt{x}-6-\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}\)
=\(\frac{x+\sqrt{x}-6}{(\sqrt{x}-3)(\sqrt{x}+3)}\)
=\(\frac{x+3\sqrt{x}-2\sqrt{x}-6}{(\sqrt{x}-3)(\sqrt{x}+3)}\)
=\(\frac{(\sqrt{x}+3)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}+3)}\)
=\(\frac{\sqrt{x}-2}{\sqrt{x}-3}\)
a)B=\(\frac{\sqrt{x}+1}{\sqrt{x}+2}=\frac{\sqrt{16}+1}{\sqrt{16}+2}=\frac{4+1}{4+2}=\frac{5}{6}\)
c)A=\(\frac{\sqrt{x}-2}{\sqrt{x}-3}=3⇒ \sqrt{x}-2=3\sqrt{x}-9⇒ 2\sqrt{x}=7\)
⇒\(\sqrt{x}=\frac{7}{2} ⇒x=\frac{49}{4}\)
