[tex] \frac{2(2 \sqrt{3} - \sqrt{2} ) 3( 2\sqrt{3}+ \sqrt{2} ) }{2+ \sqrt{2} } [/tex] jika dirasionalkan.. ²log[tex] \frac{1}{6} [/tex] - ⁵log[tex] \sqrt{7} [/t
Matematika
AzuraRaiser
Pertanyaan
[tex] \frac{2(2 \sqrt{3} - \sqrt{2} ) 3( 2\sqrt{3}+ \sqrt{2} ) }{2+ \sqrt{2} } [/tex] jika dirasionalkan..
²log[tex] \frac{1}{6} [/tex] - ⁵log[tex] \sqrt{7} [/tex] x ⁷log[tex] \frac{1}{5} [/tex] / ³log[tex]9 \sqrt{3} [/tex]
adalah...
²log[tex] \frac{1}{6} [/tex] - ⁵log[tex] \sqrt{7} [/tex] x ⁷log[tex] \frac{1}{5} [/tex] / ³log[tex]9 \sqrt{3} [/tex]
adalah...
2 Jawaban
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1. Jawaban Anonyme
Kelas 10 Matematika
Bab Akar, Pangkat dan Logaritma
1] (2 (2√3 - √2) . 3 . (2√3 + √2))/(2 + √2)
= (6 ((2√3)² - (√2)²)/(2 + √2) . (2 - √2)/(2 - √2)
= (6 (12 - 2) (2 - √2)) / (2² - √2²)
= (6 . 10 (2 - √2)) / (4 - 2)
= (120 - 60 √2)/2
= 60 - 30√2
2] (²log (1/6) - ⁵log √7 . ⁷log (1/5))/(³log 9√3)
= (²log 6^-1 - ⁵log 5 . ⁷log √7)/(³log 3^(2 1/2))
= (-1 ²log 6 - 1 . 1/2) / (5/2 . ³log 3)
= (- ²log (2 . 3) - 1/2) . 2/5
= (2/5) (-²log 2 - ²log 3 - 1/2)
= (2/5) . (-1 - ²log 3 - 1/2)
= (2/5) . (-3/2 - ²log 3) -
2. Jawaban Anonyme
Jawaban pd lampiran
Catatan :
Lbh rapi lg ngetik soal, pisahkan dg tanda kurung antara pembilang dan penyebut.
:)Pertanyaan Lainnya