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8/3-(sqrt(6))

Jul 22, 2017
edited by Guest  Jul 22, 2017

#1
+99388
+2

8/3-(sqrt(6))

I assume you mean

8/[3-sqrt(6)]

you have to multiply by the conjugate of the denominator.

$$\frac{8}{3-\sqrt6}\\ =\frac{8}{3-\sqrt6}\times \frac{3+\sqrt6}{3+\sqrt6}\\ =\frac{8(3+\sqrt6)}{9-6}\\ =\frac{24+8\sqrt6}{3}\\$$

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Jul 22, 2017
#2
+1900
0

I assume you mean $$\frac{8}{3}-\sqrt{6}$$

$$\frac{8}{3}-\sqrt{6}$$

$$\frac{8}{3}-\frac{3\sqrt{6}}{3}$$

$$\frac{8-3\times\sqrt{6}}{3}$$

$$\frac{8-3\times\sqrt{6}}{3}$$

$$\frac{8-3\times2.4494897427831781}{3}$$

$$\frac{8-7.3484692283495343}{3}$$

$$\frac{0.6515307716504657}{3}$$

$$0.2171769238834885667$$

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Jul 23, 2017