a: \(A=\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{4-x}:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2}{x}\right)\)
\(=\dfrac{8\sqrt{x}-4x+8x}{4-x}:\dfrac{x-\sqrt{x}-2\sqrt{x}+4}{x\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x\left(\sqrt{x}-2\right)}{x-3\sqrt{x}+4}\)
\(=\dfrac{-4\sqrt{x}}{x-3\sqrt{x}+4}\)
b: Để A=-2 thì \(-4\sqrt{x}=-2x+6\sqrt{x}-8\)
\(\Leftrightarrow2x-10\sqrt{x}+8=0\)
=>x=1(nhận) hoặc x=16(nhận)
Đúng 0
Bình luận (0)
