Sửa đề: \(A=\left(\dfrac{1}{2\sqrt{x}-3}-\dfrac{3}{2\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{16\sqrt{x}-21}{2x+\sqrt{x}-3}\)
a: ĐKXĐ: x>=0; \(x\notin\left\{1;\dfrac{9}{4};\dfrac{441}{256}\right\}\)
b: \(A=\left(\dfrac{2\sqrt{x}+3-6\sqrt{x}+9}{4x-9}+\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{16\sqrt{x}-21}\)
\(=\left(\dfrac{-4\sqrt{x}+12}{4x-9}+\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{16\sqrt{x}-21}\)
\(=\dfrac{-4x+4\sqrt{x}+12\sqrt{x}-12+4x-9}{\left(2\sqrt{x}+3\right)\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{16\sqrt{x}-21}\)
\(=\dfrac{16\sqrt{x}-21}{2\sqrt{x}-3}\cdot\dfrac{1}{16\sqrt{x}-21}=\dfrac{1}{2\sqrt{x}-3}\)
c: Để A<0 thì \(2\sqrt{x}-3< 0\)
=>0<x<9/4