Đáp án:
Giải thích các bước giải:
`C=(\frac{3+\sqrt{x}}{3-\sqrt{x}}-\frac{3-\sqrt{x}}{3+\sqrt{x}}-\frac{4x}{x-9}):(\frac{5}{3-\sqrt{x}}-\frac{4\sqrt{x}+2}{3\sqrt{x}-x})`
ĐK: `x>0,x \ne 9`
`C=[\frac{(3+\sqrt{x})^2}{(3+\sqrt{x})(3-\sqrt{x})}-\frac{(3-\sqrt{x})^2}{(3+\sqrt{x})(3-\sqrt{x})}+\frac{4x}{(3+\sqrt{x})(3-\sqrt{x})}]:[\frac{5\sqrt{x}}{\sqrt{x}(3-\sqrt{x})}-\frac{4\sqrt{x}+2}{\sqrt{x}(3-\sqrt{x})}]`
`C=[\frac{9+6\sqrt{x}+x}{(3+\sqrt{x})(3-\sqrt{x})}-\frac{9-6\sqrt{x}+x}{(3+\sqrt{x})(3-\sqrt{x})}+\frac{4x}{(3+\sqrt{x})(3-\sqrt{x})}]:[\frac{5\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}(3-\sqrt{x})}]`
`C=[\frac{9+6\sqrt{x}+x-9+6\sqrt{x}-x+4x}{(3+\sqrt{x})(3-\sqrt{x})}].[\frac{\sqrt{x}(3-\sqrt{x})}{\sqrt{x}-2}]`
`C=[\frac{12\sqrt{x}+4x}{(3+\sqrt{x})(3-\sqrt{x})}].[\frac{\sqrt{x}(3-\sqrt{x})}{\sqrt{x}-2}]`
`C=\frac{4\sqrt{x}(3+\sqrt{x})}{(3+\sqrt{x})(3-\sqrt{x})}.\frac{\sqrt{x}(3-\sqrt{x})}{\sqrt{x}-2}`
`C=\frac{4x}{\sqrt{x}-2}`
