Berapa hasil pengoperasian deret di bawah ini?
[tex]\displaystyle \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+2}+\cdots+\frac{1}{2\sqrt{2}+3}[/tex]
[tex]\displaystyle \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+2}+\cdots+\frac{1}{2\sqrt{2}+3}[/tex]
Verified answer
Jawab:
2
Penjelasan dengan langkah-langkah:
[tex]\displaystyle\frac{1}{1+\sqrt{2}}=\frac{1-\sqrt{2}}{1^2-(\sqrt{2})^2}=\frac{1-\sqrt{2}}{1-2}=\frac{1-\sqrt{2}}{-1}=-1+\sqrt{2}=\sqrt{2}-\sqrt{1}[/tex]
[tex]\displaystyle\frac{1}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2})^2-(\sqrt{3})^2}=\frac{\sqrt{2}-\sqrt{3}}{2-3}=\frac{\sqrt{2}-\sqrt{3}}{-1}=-\sqrt{2}+\sqrt{3}=\sqrt{3}-\sqrt{2}[/tex]
Polanya udah ketebak
[tex]\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...\\...-\sqrt{3})+(\sqrt{2}-\sqrt{2})-\sqrt{1}[/tex]
[tex]2\sqrt{2}+3=\sqrt{4(2)}+\sqrt{9}=\sqrt{8}+\sqrt{9}[/tex]
Bilangan akhir adalah
[tex]\sqrt{9}[/tex]
atau 3
Maka
[tex]...-\sqrt{3})+(\sqrt{2}-\sqrt{2})-\sqrt{1}[/tex]
[tex]=3-\sqrt{1}[/tex]
= 3 - 1
= 2
(xcvi)