Question

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Answer 1

\(P=\left(\dfrac{a}{2\sqrt{a}}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{\left(a-\sqrt{a}\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(a+\sqrt{a}\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\left(\dfrac{a-1}{2\sqrt{a}}\right)\left(\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{2\sqrt{a}}.\dfrac{-4a}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=-2\sqrt{a}\)

Để \(P\ge-2\Leftrightarrow-2\sqrt{a}\ge-2\)

\(\Leftrightarrow\sqrt{a}\le1\Rightarrow a\le1\)

Kết hợp ĐKXĐ ta được: \(0< a< 1\)