1: \(A=\dfrac{x\sqrt{x}+2x+4\sqrt{x}-\left(x-3\right)\left(\sqrt{x}-2\right)-7\sqrt{x}-10}{x\sqrt{x}-8}\cdot\dfrac{x+2\sqrt{x}+4}{\sqrt{x}+7}\)
\(=\dfrac{x\sqrt{x}+2x-3\sqrt{x}-10-x\sqrt{x}+2x+3\sqrt{x}-6}{\sqrt{x}-2}\cdot\dfrac{1}{\sqrt{x}+7}\)
\(=\dfrac{4x-16}{\sqrt{x}-2}\cdot\dfrac{1}{\sqrt{x}+7}=\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}+7}\)
Để A<2 thì \(4\sqrt{x}+8-2\sqrt{x}-14< 0\)
hay x<36
Kết hợp ĐKXĐ, ta được:
\(\left\{{}\begin{matrix}0< =x< 36\\x< >4\end{matrix}\right.\)
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