Bài 5:
Ta có: \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(\Rightarrow\sqrt{A}=\sqrt{\dfrac{\sqrt{x}-2}{\sqrt{x}+1}}\)
Mà: \(\sqrt{A}< \dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{\dfrac{\sqrt{x}-2}{\sqrt{x}+1}}< \dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< \dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{9\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)}{9\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\dfrac{9\sqrt{x}-18-\sqrt{x}-1}{9\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\dfrac{8\sqrt{x}-19}{9\left(\sqrt{x}+1\right)}< 0\Leftrightarrow8\sqrt{x}-19< 9\)
\(\Leftrightarrow8\sqrt{x}< 19\)
\(\Leftrightarrow\sqrt{x}< \dfrac{19}{8}\)
\(\Leftrightarrow x< \dfrac{361}{64}\)
Kết hợp với đk:
\(0\le x< \dfrac{361}{64}\)
Bài 6:
a) \(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(A=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(A=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b) \(A>\dfrac{1}{6}\) khi
\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}>\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{6\cdot\left(\sqrt{x}-2\right)-3\sqrt{x}}{6\cdot3\sqrt{x}}>0\)
\(\Leftrightarrow\dfrac{6\sqrt{x}-12-3\sqrt{x}}{18\sqrt{x}}>0\)
\(\Leftrightarrow\dfrac{3\sqrt{x}-12}{18\sqrt{x}}>0\)
\(\Leftrightarrow3\sqrt{x}-12>0\)
\(\Leftrightarrow3\sqrt{x}>12\)
\(\Leftrightarrow\sqrt{x}>4\)
\(\Leftrightarrow x>16\)
c) \(\sqrt{A}< \dfrac{1}{2}\)
\(\Leftrightarrow A< \dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-2}{3\sqrt{x}}< \dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4\cdot\left(\sqrt{x}-2\right)-3\sqrt{x}}{4\cdot3\sqrt{x}}< 0\)
\(\Leftrightarrow\dfrac{4\sqrt{x}-8-3\sqrt{x}}{12\sqrt{x}}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-8}{12\sqrt{x}}< 0\)
\(\Leftrightarrow\sqrt{x}-8< 0\)
\(\Leftrightarrow x< 64\)
Kết hợp với đk:
\(\Leftrightarrow0< x< 64\)
