Question

Answer 1

\(1,\\ a,=2\sqrt{3}+5\sqrt{3}-6\sqrt{3}=\sqrt{3}\\ b,=15+2\sqrt{5}-2\sqrt{5}=15\\ 2,\\ 1,ĐK:x\ge0\\ PT\Leftrightarrow7\sqrt{2x}-2\sqrt{2x}-3\sqrt{2x}=4\\ \Leftrightarrow\sqrt{2x}=2\Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\\ 2,A=\sqrt{3}-1-\left(2+\sqrt{3}\right)=-3\\ 3,\\ a,P=\dfrac{\sqrt{x}+\sqrt{x}-2-2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{2}\\ P=\dfrac{-6}{2\left(\sqrt{x}-2\right)}=-\dfrac{3}{\sqrt{x}-2}\\ b,P=-\dfrac{3}{\sqrt{x}-2}\)

Vì \(\sqrt{x}-2\ge2\Leftrightarrow-\dfrac{3}{\sqrt{x}-2}\ge-\dfrac{3}{2}\)

Vậy \(P_{min}=-\dfrac{3}{2}\Leftrightarrow x=0\)