Question

Jika [tex]\displaystyle x=1+\sqrt[5]{2}+\sqrt[5]{4}+\sqrt[5]{8}+\sqrt[5]{16}[/tex] maka nilai dari [tex]\displaystyle \left ( 1+\frac{1}{x} \right )^{-10}[/tex] adalah ...

Answer 1

Verified answer

Jawab:
[tex]\displaystyle=\bf\frac{1}{4}[/tex]

Penjelasan dengan langkah-langkah:
Jika
[tex]x=1+\sqrt[5]{2}+\sqrt[5]{4}+\sqrt[5]{8}+\sqrt[5]{16}[/tex]
maka tentukan nilai dari
[tex]\displaystyle\left(1+\frac{1}{x}\right)^{-10}[/tex]
Cari dulu nilai dari
[tex]\displaystyle\frac{1}{x}=\frac{1}{1+\sqrt[5]{2}+\sqrt[5]{4}+\sqrt[5]{8}+\sqrt[5]{16}}[/tex]
Lalu rasionalkan.
[tex]\displaystyle\frac{1}{x}=\frac{\sqrt[5]{2}-1}{(\sqrt[5]{2}-1)(1+\sqrt[5]{2}+\sqrt[5]{4}+\sqrt[5]{8}+\sqrt[5]{16})}[/tex]
Dengan sifat distributif, sederhanakan penyebut.
[tex]\displaystyle\frac{1}{x}=\frac{\sqrt[5]{2}-1}{(2^{\frac{1}{5}}-1)(1+2^{\frac{1}{5}}+4^{\frac{1}{5}}+8^{\frac{1}{5}}+16^{\frac{1}{5}})}[/tex]
[tex]\displaystyle\frac{1}{x}=\frac{\sqrt[5]{2}-1}{(2^{\frac{1}{5}}-1)(1+2^{\frac{1}{5}}+(2^2)^{\frac{1}{5}}+(2^3)^{\frac{1}{5}}+(2^4)^{\frac{1}{5}})}[/tex]
[tex]\displaystyle\frac{1}{x}=\frac{\sqrt[5]{2}-1}{(2^{\frac{1}{5}}-1)(1+2^{\frac{1}{5}}+2^{\frac{2}{5}}+2^{\frac{3}{5}}+2^{\frac{4}{5}})}[/tex]
[tex]\displaystyle\frac{1}{x}=\frac{\sqrt[5]{2}-1}{2^{\frac{1}{5}}+2^{\frac{1}{5}}(2^{\frac{1}{5}})+2^{\frac{1}{5}}(2^{\frac{2}{5}})+2^{\frac{1}{5}}(2^{\frac{3}{5}})+2^{\frac{1}{5}}(2^{\frac{4}{5}})-(1+2^{\frac{1}{5}}+2^{\frac{2}{5}}+2^{\frac{3}{5}}+2^{\frac{4}{5}})}[/tex]
Sifat eksponen yang digunakan:
[tex]\because n^a(n^b)=n^{a+b}\therefore[/tex]
[tex]\displaystyle\frac{1}{x}=\frac{\sqrt[5]{2}-1}{2^{\frac{1}{5}}+2^{\frac{1}{5}+\frac{1}{5}}+2^{\frac{1}{5}+\frac{2}{5}}+2^{\frac{1}{5}+\frac{3}{5}}+2^{\frac{1}{5}+\frac{4}{5}}-(1+2^{\frac{1}{5}}+2^{\frac{2}{5}}+2^{\frac{3}{5}}+2^{\frac{4}{5}})}[/tex]
[tex]\displaystyle\frac{1}{x}=\frac{\sqrt[5]{2}-1}{2^{\frac{1}{5}}+2^{\frac{2}{5}}+2^{\frac{3}{5}}+2^{\frac{4}{5}}+2^{\frac{5}{5}}-(1+2^{\frac{1}{5}}+2^{\frac{2}{5}}+2^{\frac{3}{5}}+2^{\frac{4}{5}})}[/tex]
[tex]\displaystyle\frac{1}{x}=\frac{\sqrt[5]{2}-1}{2^{\frac{1}{5}}+2^{\frac{2}{5}}+2^{\frac{3}{5}}+2^{\frac{4}{5}}+2^{1}-(1+2^{\frac{1}{5}}+2^{\frac{2}{5}}+2^{\frac{3}{5}}+2^{\frac{4}{5}})}[/tex]
[tex]\displaystyle\frac{1}{x}=\frac{\sqrt[5]{2}-1}{2^{\frac{1}{5}}+2^{\frac{2}{5}}+2^{\frac{3}{5}}+2^{\frac{4}{5}}+2-(1+2^{\frac{1}{5}}+2^{\frac{2}{5}}+2^{\frac{3}{5}}+2^{\frac{4}{5}})}[/tex]
[tex]\displaystyle\frac{1}{x}=\frac{\sqrt[5]{2}-1}{\not2^{\frac{1}{5}}+\not2^{\frac{2}{5}}+\not2^{\frac{3}{5}}+\not2^{\frac{4}{5}}+2-1-\not2^{\frac{1}{5}}-\not2^{\frac{2}{5}}-\not2^{\frac{3}{5}}-\not2^{\frac{4}{5}}}[/tex]
[tex]\displaystyle\frac{1}{x}=\frac{\sqrt[5]{2}-1}{1}[/tex]
[tex]\displaystyle\frac{1}{x}=\sqrt[5]{2}-1[/tex]
Oleh karena itu
[tex]\displaystyle\left(1+\frac{1}{x}\right)^{-10}[/tex]
[tex]\displaystyle=\left(\not1+\sqrt[5]{2} -\not1\right)^{-10}[/tex]
[tex]\displaystyle=\left(\sqrt[5]{2}\right)^{-10}[/tex]
Dengan sifat eksponen berikut:
[tex]\because\sqrt[r]{p^q} =p^{\frac{q}{r}}\therefore[/tex]
[tex]\displaystyle=\left(\sqrt[5]{2}\right)^{-10}=\sqrt[5]{2^{-10}}=2^{\frac{-10}{5}}=2^{-2}[/tex]
Lagi, dengan sifat eksponen berikut:
[tex]\displaystyle\because n^{-a}=\frac{1}{n^a}\therefore[/tex]
[tex]\displaystyle2^{-2}=\frac{1}{2^2}[/tex]
[tex]\displaystyle=\bf\frac{1}{4}[/tex]
(xcvi)