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\(ĐKXĐ:x\ge0;x\ne\frac{1}{9}\)
\(\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(\left(\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{9x-1}\right):\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(\frac{3x-3\sqrt{x}+\sqrt{x}-1+5\sqrt{x}+1}{9x-1}.\frac{3\sqrt{x}+1}{3}\)
\(\frac{3x+3\sqrt{x}}{9x-1}.\frac{3\sqrt{x}+1}{3}\)
\(\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{x}{\sqrt{x}-1}+\frac{\sqrt{x}-2x}{x-\sqrt{x}}\)
\(=\frac{x}{\sqrt{x}-1}+\frac{1-2\sqrt{x}}{\sqrt{x}-1}=\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
http://123link.pro/U9cCj2bj
\(\frac{2x-\sqrt{x}}{\sqrt{x}-1}+\frac{x}{\sqrt{x}-1}\)
\(=\frac{2x+x-\sqrt{x}}{\sqrt{x}-1}\)
\(=\frac{3x-\sqrt{x}}{\sqrt{x-1}}\)
P/s đề thiếu hay đủ vậy em : ))