Question

Giúp e vs

Answer 1

\(a,B=\dfrac{x+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\\ b,x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\\ x=5+\sqrt{2}-4-\sqrt{2}=1\\ \Leftrightarrow B=\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\\ c,B-\dfrac{1}{3}=\dfrac{3\sqrt{x}+3-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}=\dfrac{2\sqrt{x}}{3\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow B>\dfrac{1}{3}\)