ĐKXĐ: x>4
A= \(\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
A= \(\left[\frac{2\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}-\frac{5\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\right]\cdot\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
A= \(\left[\frac{4\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}+\frac{3\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}-\frac{5\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\right]\)\(\cdot\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
A= \(\frac{2\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\cdot\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)\(=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)\(=2,5-\frac{2,5}{2\sqrt{x}+1}\)
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