Berapa nilai dari penyerderhanaan bentuk dibawah ini?
[tex]\displaystyle 1+\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}[/tex]
[tex]\displaystyle 1+\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}[/tex]
Verified answer
Jawab:
[tex]=\sqrt{2}[/tex]
Penjelasan:
Sederhanakanlah
[tex]\displaystyle 1+\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}=\\\\1+\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+\sqrt{2\cdot3}+\sqrt{4\cdot2}+4}=\\\\1+\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+\sqrt{2}\sqrt{3}+\sqrt{4}\sqrt{2}+(2+2)}=\\\\1+\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+\sqrt{2}\sqrt{3}+2\sqrt{2}+(\sqrt{2}\sqrt{2}+2)}=[/tex]
[tex]\displaystyle1+\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2+(\sqrt{2}\sqrt{2}+\sqrt{2}\sqrt{3}+2\sqrt{2})}=\\\\1+\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2+\sqrt{2}(\sqrt{2}+\sqrt{3}+2)}=\\\\1+\frac{\sqrt{2}+\sqrt{3}+2}{1(\sqrt{2}+\sqrt{3}+2)+\sqrt{2}(\sqrt{2}+\sqrt{3}+2)}=\\\\1+\frac{\not\sqrt{2}+\not\sqrt{3}+\not2}{(1+\sqrt{2})(\not\sqrt{2}+\not\sqrt{3}+\not2)}=\\\\1+\frac{1}{1+\sqrt{2}}=\\\\\frac{2+\sqrt{2}}{1+\sqrt{2}}=\frac{(2+\sqrt{2})(1-\sqrt{2})}{1^2-(\sqrt{2})^2}[/tex]
[tex]\displaystyle=\frac{2-2\sqrt{2}+\sqrt{2}-2}{1-2}\\\\=\frac{-\sqrt{2}}{-1}[/tex]
[tex]\displaystyle=\sqrt{2}[/tex]
(xcvi)