Chỗ biểu thức A mình sửa thành \(A=\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\) nha
\(a,x=16\Leftrightarrow B=\dfrac{4+3}{16+4+1}=\dfrac{7}{21}=\dfrac{1}{3}\\ b,P=A:B=\dfrac{2x+1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+3}\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\\ c,P< \dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{1}{2}< 0\\ \Leftrightarrow\dfrac{2\sqrt{x}-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\\ \Leftrightarrow\dfrac{\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\\ \Leftrightarrow\sqrt{x}-3< 0\left(\sqrt{x}+3\ge3>0\right)\\ \Leftrightarrow x< 9\\ \Leftrightarrow0\le x< 9\)
