Dengan cara merasionalkan bagian penyebutnya, pecahan 12+√18 per √6 ekuivalen dengan...
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Pakai cara ya.. Tks
MicoArrafi
5 votes
Thanks 9
miyatun299
Kenapa hasilny 2√6+√3 dan knp
izharulhaq78
![\displaystyle \frac{12+\sqrt{18}}{\sqrt6}=\frac{12+\sqrt{9\times2}}{\sqrt6}\times\frac{\sqrt6}{\sqrt6}\\\frac{12+\sqrt{18}}{\sqrt6}=\frac{[12+3\sqrt{2}]\sqrt6}{(\sqrt6)^2}\\\frac{12+\sqrt{18}}{\sqrt6}=\frac{3[4+\sqrt{2}]\sqrt6}{6}\\\frac{12+\sqrt{18}}{\sqrt6}=\frac{4\sqrt6+\sqrt{12}}{2}\\\frac{12+\sqrt{18}}{\sqrt6}=\frac{4\sqrt6+\sqrt{4\times3}}{2}\\\frac{12+\sqrt{18}}{\sqrt6}=\frac{4\sqrt6+2\sqrt{3}}{2}\\\frac{12+\sqrt{18}}{\sqrt6}=\frac{2[2\sqrt6+\sqrt{3}]}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cfrac%7B12%2B%5Csqrt%7B18%7D%7D%7B%5Csqrt6%7D%3D%5Cfrac%7B12%2B%5Csqrt%7B9%5Ctimes2%7D%7D%7B%5Csqrt6%7D%5Ctimes%5Cfrac%7B%5Csqrt6%7D%7B%5Csqrt6%7D%5C%5C%5Cfrac%7B12%2B%5Csqrt%7B18%7D%7D%7B%5Csqrt6%7D%3D%5Cfrac%7B%5B12%2B3%5Csqrt%7B2%7D%5D%5Csqrt6%7D%7B%28%5Csqrt6%29%5E2%7D%5C%5C%5Cfrac%7B12%2B%5Csqrt%7B18%7D%7D%7B%5Csqrt6%7D%3D%5Cfrac%7B3%5B4%2B%5Csqrt%7B2%7D%5D%5Csqrt6%7D%7B6%7D%5C%5C%5Cfrac%7B12%2B%5Csqrt%7B18%7D%7D%7B%5Csqrt6%7D%3D%5Cfrac%7B4%5Csqrt6%2B%5Csqrt%7B12%7D%7D%7B2%7D%5C%5C%5Cfrac%7B12%2B%5Csqrt%7B18%7D%7D%7B%5Csqrt6%7D%3D%5Cfrac%7B4%5Csqrt6%2B%5Csqrt%7B4%5Ctimes3%7D%7D%7B2%7D%5C%5C%5Cfrac%7B12%2B%5Csqrt%7B18%7D%7D%7B%5Csqrt6%7D%3D%5Cfrac%7B4%5Csqrt6%2B2%5Csqrt%7B3%7D%7D%7B2%7D%5C%5C%5Cfrac%7B12%2B%5Csqrt%7B18%7D%7D%7B%5Csqrt6%7D%3D%5Cfrac%7B2%5B2%5Csqrt6%2B%5Csqrt%7B3%7D%5D%7D%7B2%7D "\displaystyle \frac{12+\sqrt{18}}{\sqrt6}=\frac{12+\sqrt{9\times2}}{\sqrt6}\times\frac{\sqrt6}{\sqrt6}\\\frac{12+\sqrt{18}}{\sqrt6}=\frac{[12+3\sqrt{2}]\sqrt6}{(\sqrt6)^2}\\\frac{12+\sqrt{18}}{\sqrt6}=\frac{3[4+\sqrt{2}]\sqrt6}{6}\\\frac{12+\sqrt{18}}{\sqrt6}=\frac{4\sqrt6+\sqrt{12}}{2}\\\frac{12+\sqrt{18}}{\sqrt6}=\frac{4\sqrt6+\sqrt{4\times3}}{2}\\\frac{12+\sqrt{18}}{\sqrt6}=\frac{4\sqrt6+2\sqrt{3}}{2}\\\frac{12+\sqrt{18}}{\sqrt6}=\frac{2[2\sqrt6+\sqrt{3}]}{2}")
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