Berapa hasil dari pengoperasian bentuk di bawah ini?
[tex]\displaystyle \sqrt{6+2\sqrt{3}+2\sqrt{2}+2\sqrt{6}}-\frac{1}{\sqrt{5-2\sqrt{6}}}[/tex]
[tex]\displaystyle \sqrt{6+2\sqrt{3}+2\sqrt{2}+2\sqrt{6}}-\frac{1}{\sqrt{5-2\sqrt{6}}}[/tex]
Verified answer
Jawab:
= 1
Penjelasan dengan langkah-langkah:
Suku kiri
[tex]\displaystyle\sqrt{6+2\sqrt{3}+2\sqrt{2}+2\sqrt{6}}\\=\sqrt{6+2\sqrt{3}+2\sqrt{2}+2\sqrt{3}\sqrt{2}}\\=\sqrt{6+2(\sqrt{3}+\sqrt{2}+\sqrt{3}\sqrt{2})}\\=\sqrt{(1+3+2)+2(1\sqrt{3}+1\sqrt{2}+\sqrt{3}\sqrt{2})}\\=\sqrt{((\sqrt{1})^2+(\sqrt{3})^2+(\sqrt{2})^2)+2(\sqrt{1}\sqrt{3}+\sqrt{1}\sqrt{2}+\sqrt{3}\sqrt{2})}[/tex]
Udah keliatan polanya
[tex]\displaystyle\because(a+b+c)^2=(a^2+b^2+c^2)+2(ab+ac+bc)\therefore\\\sqrt{((\sqrt{1})^2+(\sqrt{3})^2+(\sqrt{2})^2)+2(\sqrt{1}\sqrt{3}+\sqrt{1}\sqrt{2}+\sqrt{3}\sqrt{2})}\\=\sqrt{(\sqrt{1}+\sqrt{3}+\sqrt{2})^2}=\sqrt{1}+\sqrt{3}+\sqrt{2}=\bf1+\sqrt{3}+\sqrt{2}[/tex]
Sekarang suku kanan
[tex]\displaystyle\frac{1}{\sqrt{5-2\sqrt{6}}}=\sqrt{5+2\sqrt{6}}\\\\\because \sqrt{p+q\pm2\sqrt{pq}}=\sqrt{p}\pm\sqrt{q}\:\:\{p > q\}\therefore\\\\\sqrt{5+2\sqrt{6}}=\sqrt{p}+\sqrt{q}\\p+q=5,\:\:\:pq=6\\3+2=5,\:\:\:3(2)=6\\\sqrt{5+2\sqrt{6}}=\bf\sqrt{3}+\sqrt{2}[/tex]
Oleh karena itu
[tex]\displaystyle\sqrt{6+2\sqrt{3}+2\sqrt{2}+2\sqrt{6}}\:\:-\:\:\frac{1}{\sqrt{5-2\sqrt{6}}}\\\\=\bf(1+\sqrt{3}+\sqrt{2})-(\sqrt{3}+\sqrt{2})\\\\=\rm1+\sqrt{3}+\sqrt{2}-\sqrt{3}-\sqrt{2}[/tex]
= 1
(xcvi)