Question

Giúp mình b2 b3 với ạ

Answer 1

\(2,\\ 1,=20\sqrt{3}+20\sqrt{3}+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=40\sqrt{3}+\sqrt{3}=41\sqrt{3}\\ 2,A=\dfrac{2\sqrt{x}-9-x+9+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{2\sqrt{x}-x+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ c,A< 1\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\left(4>0\right)\\ \Leftrightarrow x< 9\Leftrightarrow0\le x< 9\)

 

Answer 2

\(3,\\ 1,A=\sqrt{2}-1-\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}=\sqrt{2}-1-\sqrt{2}=-1\\ 2,\\ a,P=\dfrac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{4}\left(x\ge0;x\ne4\right)\\ P=\dfrac{4\left(\sqrt{x}+2\right)}{4\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\\ b,P< 1\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-1< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-2}< 0\Leftrightarrow\sqrt{x}-2< 0\left(4>0\right)\\ \Leftrightarrow x< 4\Leftrightarrow0\le x< 4\)