[tex]\it \Big[\dfrac{1}{\sqrt{10}-3}\Big]=6\\ \\ \\ \Big[\dfrac{1}{\sqrt{10}}-3\Big]=-3[/tex]
[tex]\it \dfrac{^{\sqrt{10}+3)}1}{\ \ \sqrt{10}-3}=\dfrac{\sqrt{10}+3}{(\sqrt{10})^2-3^2}=\dfrac{\sqrt{10}+3}{10-9}=\sqrt{10}+3>\sqrt9+3=3+3=6\\ \\ \\[/tex]
[tex]\it 9<10<16 \Rightarrow \sqrt9<\sqrt{10}<\sqrt{16} \Rightarrow 3<\sqrt{10}<4|_{+3} \Rightarrow 6<\sqrt{10}+3<7[/tex]
