a) ĐKXĐ: \(x\ne4\)và \(x>0\)
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\(\Leftrightarrow A=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{6}{3\left(\sqrt{x}-2\right)}+\frac{1}{\sqrt{x}+2}\right)\)\(:\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{3x-6\sqrt{x}\left(\sqrt{x}+2\right)+3\sqrt{x}\left(\sqrt{x}-2\right)}{3\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}:\left(\frac{x-2+10-x}{\sqrt{x}-2}\right)\)
\(\Leftrightarrow A=\frac{3x-6x-12\sqrt{x}+3x-6\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\left(\frac{8}{\sqrt{x}-2}\right)\)
\(\Leftrightarrow A=\frac{-18\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}-2}{8}\)
\(\Leftrightarrow A=\frac{-3}{4\left(\sqrt{x}+2\right)}\)
Vậy \(A=\frac{-3}{4\left(\sqrt{x}-2\right)}\)với \(x>0\)và \(x\ne4\)
b)Ta có \(A< 2\Leftrightarrow\frac{-3}{4\left(\sqrt{x}-2\right)}< 2\)
\(\Leftrightarrow\frac{-3}{4\left(\sqrt{x}-2\right)}-2< 0\)
\(\Leftrightarrow\frac{-3-8\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow\frac{-3-8\sqrt{x}-16}{4\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow\frac{-18-8\sqrt{x}}{4\left(\sqrt{x}-2\right)}< 0\)
\(\Leftrightarrow-18-8\sqrt{x}< 0\)( Vì \(4\left(\sqrt{x}-2\right)>0\)với \(\forall x\))
\(\Leftrightarrow\sqrt{x}< \frac{-9}{4}\)(Vô Nghiệm)
Vậy không có gtr nào của x thỏa mãn A<2
Mình làm nhầm bạn ơi, bỏ câu trả lời ý đi nha