tolong bantu jawab nomer 8 dan 9 dong kak...

~Mapel : Matematika~
`Bab : Eksponensial~

8)

[tex]( \frac{16}{81} ) ^{ \frac{ - 3}{4} } [/tex]
= ((2^4)/3^4 ))^-3/4

= (2^-3/3^-3)
= 3^3/2^3
= 27/8


9)

[tex] \sqrt{48} \div \sqrt{6} \times \sqrt{12} [/tex]


[tex] = \sqrt{8} \times \sqrt{12} [/tex]
[tex] = \sqrt{96} [/tex]


[tex] = \sqrt{16 \times 6} [/tex]


[tex] = 4 \sqrt{6} [/tex]

Mapel : Matematika
Materi : Akar dan Pangkat

Soal No. 8

[tex]=( \frac{16}{81})^{- \frac{3}{4} } \\ \\ =( \frac{2^4}{3^4})^{- \frac{3}{4} } \\ \\ =(2^4\times3^{-4}})^{- \frac{3}{4}}\\\\ = 2^{(4\times(- \frac{3}{4} ))}\times3^{((-4)\times(- \frac{3}{4} ))}}\\ \\=2^{-3}\times3^{3} \\ \\ = \frac{3^{3}}{2^{3}} \\ \\ = \frac{27}{8} [/tex]

Soal No. 9
Jika dalam satu kalimat pertanyaan terdapat operasi (÷) dan (×), maka selesaiakan operasi tersebut dari paling kiri.

[tex]=\frac{\sqrt{48}}{\sqrt{8}}\times \sqrt{12}\\\\=\frac{\sqrt{16\times3}}{\sqrt{4\times2}}\times\sqrt{4\times3}\\\\=\frac{4\sqrt{3}}{2\sqrt{2}}\times2\sqrt{3}\\\\=\frac{(2\sqrt{3})^2}{\sqrt{2}}\\\\=\frac{4\times3}{\sqrt{2}}\\\\=\frac{12}{\sqrt{2}}............dikalikan\ akar\ sekawan\\\\=\frac{12}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}} \\ \\=\frac{12\sqrt{2}}{(\sqrt{2})^2}\\ \\=\frac{12\sqrt{2}}{2} \\ \\ =6\sqrt{2} [/tex]

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