Question

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Answer 1

Jawab:
[tex]\displaystyle(26)=9\sqrt[12]{3}\\(27)=2^{2x-2}[/tex]

Penjelasan dengan langkah-langkah:
[tex]\displaystyle(26)\:\:\sqrt{27\sqrt{3\sqrt[3]{81} }}=\sqrt{27\sqrt{3\sqrt[3]{3^4} }}=\sqrt{3^3\sqrt{3(3)^{\frac{4}{3}}}}\\\\=\sqrt{3^3(3(3)^{\frac{4}{3}})^\frac{1}{2}}=\sqrt{3^3(3^{1+\frac{4}{3}})^\frac{1}{2}}=\sqrt{3^3(3^{\frac{3}{3}+\frac{4}{3}})^\frac{1}{2}}\\\\=\sqrt{3^3(3^{\frac{7}{3}})^\frac{1}{2}}=\sqrt{3^3(3^{\frac{7}{6}})}=\sqrt{3^{3+\frac{7}{6}}}=\sqrt{3^{\frac{18}{6}+\frac{7}{6}}}\\\\=\sqrt{3^{\frac{25}{6}}}=(3^{\frac{25}{6}})^{\frac{1}{2}}=3^{\frac{25}{12}}=3^{\frac{24+1}{12}}[/tex]

[tex]\displaystyle=3^{\frac{24}{12}}\cdot 3^{\frac{1}{12}}=3^2\cdot \sqrt[12]{3}=9\sqrt[12]{3}[/tex]
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[tex]\displaystyle(29)\:\:\sqrt{\frac{8^{2x-\frac{8}{3}}}{4^{x-2}}}=\sqrt{\frac{(2^3)^{2x-\frac{8}{3}}}{(2^2)^{x-2}}}=\left(\frac{(2^3)^{2x-\frac{8}{3}}}{(2^2)^{x-2}}\right)^{\frac{1}{2}}\\\\=\left(\frac{2^{3(2x-\frac{8}{3})}}{2^{2(x-2)}}\right)^{\frac{1}{2}}=\left(\frac{2^{6x-8}}{2^{2x-4}}\right)^{\frac{1}{2}}=\frac{2^{\frac{1}{2}(6x-8)}}{2^{\frac{1}{2}(2x-4)}}\\\\=\frac{2^{3x-4}}{2^{x-2}}=2^{3x-4-(x-2)}=2^{3x-4-x+2}=2^{2x-2}[/tex]

(xcvi)