Answer :

[tex]\frac{8}{2-\sqrt{12}}=\frac{8}{2-\sqrt{4\cdot3}}=\frac{8}{2-2\sqrt3}=\frac{8}{2(1-\sqrt3)}=\frac{4}{1-\sqrt3}\\\\\frac{4}{1-\sqrt3}\cdot\frac{1+\sqrt3}{1+\sqrt3}=\frac{4(1+\sqrt3)}{1^2-(\sqrt3)^2}=\frac{4(1+\sqrt3)}{1-3}=\frac{4(1+\sqrt3)}{-2}=-2(1+\sqrt3)}\\\\=-2-2\sqrt3[/tex]



[tex]if\ \frac{8}{2}-\sqrt{12}=4-\sqrt{4\cdot3}=4-2\sqrt3[/tex]
konrad509
[tex]\frac{8}{2-\sqrt{12}}\\ \frac{8(2+\sqrt{12})}{4-12}=\\ \frac{8(2+\sqrt{12})}{-8}=\\ -(2+\sqrt{12}})=\\ -2-\sqrt{12}=\\-2-2\sqrt3 [/tex]

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