[tex]\it a=\Big(\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt3}\Big)\cdot\sqrt6-\sqrt3=\sqrt3+\sqrt2-\sqet3=\sqrt2\\ \\ \\ b=\dfrac{2\sqrt2-2\sqrt3}{\sqrt3-\sqrt2}+2+\sqrt3=\dfrac{-2(\sqrt3-\sqrt2)}{\sqrt3-\sqrt2}+2+\sqrt3=\sqrt3[/tex]
[tex]\it \dfrac{6}{5}<\dfrac{b}{a}<\dfrac{5}{4} \Rightarrow 1,2<\dfrac{\sqrt3}{\sqrt2}<1,25 \Rightarrow 1,2^2<\Big(\dfrac{\sqrt3}{\sqrt2}\Big)^2<1,25^2 \Rightarrow\\ \\ \\ \Rightarrow 1,44<\dfrac{3}{2}<1,5625 \Rightarrow 1,44<1,5<1,5625\ \ (A)[/tex]
