\(a,A=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ b,A=\dfrac{\sqrt{a}-1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}\)
Ta có \(a>0\Leftrightarrow\dfrac{1}{\sqrt{a}}>0\Leftrightarrow-\dfrac{1}{\sqrt{a}}< 0\Leftrightarrow A=1-\dfrac{1}{\sqrt{a}}< 1-0=1\left(đpcm\right)\)
\(c,A=\dfrac{1}{2}\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\Leftrightarrow\sqrt{a}=2\Leftrightarrow a=4\left(tm\right)\)
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