Question

giups minh vs a

Answer 1

\(a.\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}\sqrt{5}-\sqrt{3}\sqrt{2}}{\sqrt{7}\sqrt{5}-\sqrt{7}\sqrt{2}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\dfrac{3}{7}}\\ b.\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}=\dfrac{\sqrt{2}\sqrt{5}+\sqrt{3}\sqrt{5}}{\sqrt{2}\sqrt{4}+\sqrt{3}\sqrt{4}}\\ =\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{4}+\left(\sqrt{2}+\sqrt{3}\right)}=\sqrt{\dfrac{5}{4}}\)

\(c.\dfrac{\sqrt{6}-2}{\sqrt{3}-\sqrt{2}}=\dfrac{\left(\sqrt{6}-2\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}\\ =\dfrac{\sqrt{18}+\sqrt{12}-2\sqrt{3}-2\sqrt{2}}{3-2}\\ =3\sqrt{2}+2\sqrt{3}-2\sqrt{3}-2\sqrt{2}=\sqrt{2}\)

\(d.\dfrac{\sqrt{24}-\sqrt{3}}{\sqrt{40}-\sqrt{5}}=\dfrac{\sqrt{3}\sqrt{8}-\sqrt{3}}{\sqrt{5}\sqrt{8}-\sqrt{5}}\\ =\dfrac{\sqrt{3}\left(\sqrt{8}-1\right)}{\sqrt{5}\left(\sqrt{8}-1\right)}=\sqrt{\dfrac{3}{5}}\)