Nilai dari(√8)pangkat 1/3???? akarnya ada di dalam kurung ya....
Reyznha
![(\sqrt{8})^{ \frac{1}{3}} = (8^{ \frac{1}{2}})^{ \frac{1}{3}} = 8^{ \frac{1}{6} } = \sqrt[6]{8} = 2^{ \frac{3}{6}} = 2^{ \frac{1}{2}} = \sqrt{2} \\\\ \sqrt{(8)^{ \frac{1}{3}}} = ( \sqrt{(2^3))^{ \frac{1}{3}}} = \sqrt{2^{ \frac{3}{3}}} = \sqrt{2^1} = \sqrt{2}](https://tex.z-dn.net/?f=+%28%5Csqrt%7B8%7D%29%5E%7B+%5Cfrac%7B1%7D%7B3%7D%7D+%3D+%288%5E%7B+%5Cfrac%7B1%7D%7B2%7D%7D%29%5E%7B+%5Cfrac%7B1%7D%7B3%7D%7D+%3D+8%5E%7B+%5Cfrac%7B1%7D%7B6%7D+%7D+%3D+%5Csqrt%5B6%5D%7B8%7D+%3D+2%5E%7B+%5Cfrac%7B3%7D%7B6%7D%7D+%3D+2%5E%7B+%5Cfrac%7B1%7D%7B2%7D%7D+%3D+%5Csqrt%7B2%7D+%5C%5C%5C%5C+%5Csqrt%7B%288%29%5E%7B+%5Cfrac%7B1%7D%7B3%7D%7D%7D+%3D+%28+%5Csqrt%7B%282%5E3%29%29%5E%7B+%5Cfrac%7B1%7D%7B3%7D%7D%7D+%3D+%5Csqrt%7B2%5E%7B+%5Cfrac%7B3%7D%7B3%7D%7D%7D+%3D+%5Csqrt%7B2%5E1%7D+%3D+%5Csqrt%7B2%7D "(\sqrt{8})^{ \frac{1}{3}} = (8^{ \frac{1}{2}})^{ \frac{1}{3}} = 8^{ \frac{1}{6} } = \sqrt[6]{8} = 2^{ \frac{3}{6}} = 2^{ \frac{1}{2}} = \sqrt{2} \\\\ \sqrt{(8)^{ \frac{1}{3}}} = ( \sqrt{(2^3))^{ \frac{1}{3}}} = \sqrt{2^{ \frac{3}{3}}} = \sqrt{2^1} = \sqrt{2}")
4 votes
Thanks 11
okysyahputra
gmnya krng mengerti tlong di uaikan saja....udh pas tp kurang mengerti
Reyznha
Bagian mana yang kurang dimengerti?
okysyahputra
ngk ad....tp nmr satu sma dgn nmr du........atau lanjutan nmr 1
okysyahputra
ok....dah mgrti.mksh ya teman..................
Reyznha
Ya.. sama-sama
Reyznha
Itu 2 cara.. :D
AghistaHaryogi
√8 1/3 -> √2 pangkat 3 dikalikan 1/3 kemudian dicoret yang 3 dan 1/3 hasilnya √2
2 votes
Thanks 3