Salut, quelqu'un peut m' aider s' il vous plaît? Ecris les quotients suivants sans radicale au dénominateur. a. -1 / V2 b. -4v3 / V6 c. 2V6 / 3V8

Bonsoir Wwwcatia

a. -1 / V2

[tex]\dfrac{-1}{\sqrt{2}}=\dfrac{(-1)\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}\\\\\dfrac{-1}{\sqrt{2}}=\dfrac{-\sqrt{2}}{(\sqrt{2})^2}\\\\\boxed{\dfrac{-1}{\sqrt{2}}=\dfrac{-\sqrt{2}}{2}}[/tex]

 b. -4v3 / V6

[tex] \dfrac{-4\sqrt{3}}{\sqrt{6}}= \dfrac{-4\sqrt{3}}{\sqrt{2}\times\sqrt{3}}\\\\\dfrac{-4\sqrt{3}}{\sqrt{6}}= \dfrac{-4}{\sqrt{2}}\\\\\dfrac{-4\sqrt{3}}{\sqrt{6}}= \dfrac{-4\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}\\\\\dfrac{-4\sqrt{3}}{\sqrt{6}}= \dfrac{-4\times\sqrt{2}}{2}\\\\\boxed{\dfrac{-4\sqrt{3}}{\sqrt{6}}= -2\sqrt{2}}[/tex]

 c. 2V6 / 3V8

[tex]\dfrac{2\sqrt{6}}{3\sqrt{8}}=\dfrac{2\sqrt{6}\times\sqrt{8}}{3\sqrt{8}\times\sqrt{8}}\\\\\dfrac{2\sqrt{6}}{3\sqrt{8}}=\dfrac{2\sqrt{48}}{3(\sqrt{8})^2}\\\\\dfrac{2\sqrt{6}}{3\sqrt{8}}=\dfrac{2\sqrt{16\times3}}{3\times8}\\\\\dfrac{2\sqrt{6}}{3\sqrt{8}}=\dfrac{2\times4\sqrt{3}}{24}\\\\\dfrac{2\sqrt{6}}{3\sqrt{8}}=\dfrac{8\sqrt{3}}{24}\\\\\boxed{\dfrac{2\sqrt{6}}{3\sqrt{8}}=\dfrac{\sqrt{3}}{3}}[/tex]

ou encore

[tex]\dfrac{2\sqrt{6}}{3\sqrt{8}}=\dfrac{2\sqrt{2\times3}}{3\sqrt{4\times2}}\\\\\dfrac{2\sqrt{6}}{3\sqrt{8}}=\dfrac{2\times\sqrt{2}\times\sqrt{3}}{3\times2\times\sqrt{2}}\\\\\dfrac{2\sqrt{6}}{3\sqrt{8}}=\dfrac{2\times\sqrt{3}}{3\times2}\\\\\boxed{\dfrac{2\sqrt{6}}{3\sqrt{8}}=\dfrac{\sqrt{3}}{3}}[/tex]

d. v7xv6 / v2xv3

[tex]\dfrac{\sqrt{7}\times\sqrt{6}}{\sqrt{2}\times\sqrt{3}}=\dfrac{\sqrt{7}\times\sqrt{6}}{\sqrt{6}}\\\\\boxed{\dfrac{\sqrt{7}\times\sqrt{6}}{\sqrt{2}\times\sqrt{3}}=\sqrt{7}}[/tex]