a) ĐKXĐ: \(x\ge0,x\ne9,x\ne4\)
b) C= \(\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
=\(\frac{x-9-x+3\sqrt{x}}{x-9}:\left(\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(3+\sqrt{x}\right)}\right)\)
= \(\frac{3\sqrt{x}-9}{x-9}:\frac{9-x-x+4\sqrt{x}-4-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
= \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-2\right)^2}\)
= \(\frac{-3}{\sqrt{x}-2}\)
Vậy C= \(\frac{-3}{\sqrt{x}-2}\)
c) Ta có C=4 =>\(\frac{-3}{\sqrt{x}-2}=4\)
\(\Leftrightarrow-3=4\sqrt{x}-8\)
\(\Leftrightarrow4\sqrt{x}=5\)
\(\Leftrightarrow\sqrt{x}=\frac{5}{4}\)
\(\Leftrightarrow x=\frac{25}{16}\left(tmđk\right)\)
Vậy với x= \(\frac{25}{16}\) thì C=4
