Question

trục mẫu

a, \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}\)

b, \(\sqrt{\dfrac{x}{y^3}+\dfrac{x}{y^4}}\)

c, \(\dfrac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)

Answer 1

a) \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)

b) \(\sqrt{\dfrac{x}{y^3}+\dfrac{x}{y^4}}=\sqrt{\dfrac{xy^4+xy^3}{y^7}}=\sqrt{\dfrac{xy^5+xy^4}{y^8}}=\dfrac{\sqrt{xy^5+xy^4}}{y^4}\)

c) \(\dfrac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}\)

Answer 2

a) \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{4.3}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}\)\(=\dfrac{\sqrt{6}}{2}\)

c)\(\dfrac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}\)