Jawab:
[tex]\displaystyle=\sf\frac{3\sqrt{9-3\sqrt5}+\sqrt{45-15\sqrt5} -2\sqrt5-2}{4}[/tex]
Penjelasan dengan langkah-langkah:
Eksponen
sifat akar :
[tex]\boxed{\sf\sqrt{(a+ b)\pm2\sqrt{ab}}=\sqrt{a}\pm\sqrt{b}} \\\\\sf\:dengan\:a > b[/tex]
soal :
Merasionalkan bentuk pecahan akar
[tex]\displaystyle\sf\frac{\sqrt{5-2\sqrt6}}{\sqrt{3-\sqrt5}}[/tex]
jawab :
Langkah 1 (pemfaktoran) :
[tex]\begin{aligned}\displaystyle\sf\frac{\sqrt{5-2\sqrt6}}{\sqrt{3-\sqrt5}}&\to\sf\frac{\sqrt{(3+2)-2\sqrt{3\times2}}}{\sqrt{3-\sqrt5}}\\&=\sf\frac{\sqrt{(\sqrt3-\sqrt2)^2}}{\sqrt{3-\sqrt5}}\\&=\sf\frac{\sqrt3 - \sqrt2}{\sqrt{3-\sqrt5}}\end{aligned}[/tex]
Langkah 2 (Rasionalkan) :
[tex]\begin{aligned}\displaystyle&=\sf\frac{\sqrt3 - \sqrt2}{\sqrt{3-\sqrt5}}\times\frac{\sqrt{3-\sqrt5}}{\sqrt{3-\sqrt5}}\\&=\sf\frac{\sqrt{3(3-\sqrt5}-\sqrt{2(3-\sqrt5)}}{(\sqrt{3-\sqrt5})(\sqrt{3-\sqrt5})}\\&=\sf\frac{\sqrt{9-3\sqrt5}-\sqrt{6-2\sqrt5}}{3-\sqrt5}\\&=\sf\frac{\sqrt{9-3\sqrt5}-\sqrt{(5+1)-2\sqrt{5\times1}}}{3-\sqrt5}\\\end{aligned}[/tex]
[tex]\begin{aligned}&=\sf\frac{\sqrt{9-3\sqrt5}-\sqrt{(\sqrt5 - \sqrt1)^2}}{3-\sqrt5}\\&=\sf\frac{\sqrt{9-3\sqrt5}-(\sqrt5 - 1)}{3-\sqrt5}\\&=\sf\frac{\sqrt{9-3\sqrt5}-\sqrt5 + 1}{3-\sqrt5}\end{aligned}[/tex]
Langkah 3 (Rasionalkan lagi) :
[tex]\begin{aligned}&=\sf\frac{\sqrt{9-3\sqrt5}-\sqrt5 + 1}{3-\sqrt5}\times\frac{3+\sqrt5}{3+\sqrt5}\\&=\sf\frac{(\sqrt{9-3\sqrt5}-\sqrt5 + 1)\times(3+\sqrt5)}{9-5}\\&=\sf\frac{3\sqrt{9-3\sqrt5}+\sqrt{5(9-3\sqrt5})-3\sqrt5-5+3+\sqrt5}{4}\\&=\sf\frac{3\sqrt{9-3\sqrt5}+\sqrt{45-15\sqrt5} -2\sqrt5-2}{4}\end{aligned}[/tex]
Kesimpulan jawaban :
∴ Hasil rasional dari [tex]\displaystyle\sf\frac{\sqrt{5-2\sqrt6}}{\sqrt{3-\sqrt5}}[/tex] adalah
Pelajari lebih lanjut : Materi Rasional Bentuk akar
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Jawab:
[tex]\displaystyle=\sf\frac{3\sqrt{9-3\sqrt5}+\sqrt{45-15\sqrt5} -2\sqrt5-2}{4}[/tex]
Penjelasan dengan langkah-langkah:
Eksponen
sifat akar :
[tex]\boxed{\sf\sqrt{(a+ b)\pm2\sqrt{ab}}=\sqrt{a}\pm\sqrt{b}} \\\\\sf\:dengan\:a > b[/tex]
soal :
Merasionalkan bentuk pecahan akar
[tex]\displaystyle\sf\frac{\sqrt{5-2\sqrt6}}{\sqrt{3-\sqrt5}}[/tex]
jawab :
Langkah 1 (pemfaktoran) :
[tex]\begin{aligned}\displaystyle\sf\frac{\sqrt{5-2\sqrt6}}{\sqrt{3-\sqrt5}}&\to\sf\frac{\sqrt{(3+2)-2\sqrt{3\times2}}}{\sqrt{3-\sqrt5}}\\&=\sf\frac{\sqrt{(\sqrt3-\sqrt2)^2}}{\sqrt{3-\sqrt5}}\\&=\sf\frac{\sqrt3 - \sqrt2}{\sqrt{3-\sqrt5}}\end{aligned}[/tex]
Langkah 2 (Rasionalkan) :
[tex]\begin{aligned}\displaystyle&=\sf\frac{\sqrt3 - \sqrt2}{\sqrt{3-\sqrt5}}\times\frac{\sqrt{3-\sqrt5}}{\sqrt{3-\sqrt5}}\\&=\sf\frac{\sqrt{3(3-\sqrt5}-\sqrt{2(3-\sqrt5)}}{(\sqrt{3-\sqrt5})(\sqrt{3-\sqrt5})}\\&=\sf\frac{\sqrt{9-3\sqrt5}-\sqrt{6-2\sqrt5}}{3-\sqrt5}\\&=\sf\frac{\sqrt{9-3\sqrt5}-\sqrt{(5+1)-2\sqrt{5\times1}}}{3-\sqrt5}\\\end{aligned}[/tex]
[tex]\begin{aligned}&=\sf\frac{\sqrt{9-3\sqrt5}-\sqrt{(\sqrt5 - \sqrt1)^2}}{3-\sqrt5}\\&=\sf\frac{\sqrt{9-3\sqrt5}-(\sqrt5 - 1)}{3-\sqrt5}\\&=\sf\frac{\sqrt{9-3\sqrt5}-\sqrt5 + 1}{3-\sqrt5}\end{aligned}[/tex]
Langkah 3 (Rasionalkan lagi) :
[tex]\begin{aligned}&=\sf\frac{\sqrt{9-3\sqrt5}-\sqrt5 + 1}{3-\sqrt5}\times\frac{3+\sqrt5}{3+\sqrt5}\\&=\sf\frac{(\sqrt{9-3\sqrt5}-\sqrt5 + 1)\times(3+\sqrt5)}{9-5}\\&=\sf\frac{3\sqrt{9-3\sqrt5}+\sqrt{5(9-3\sqrt5})-3\sqrt5-5+3+\sqrt5}{4}\\&=\sf\frac{3\sqrt{9-3\sqrt5}+\sqrt{45-15\sqrt5} -2\sqrt5-2}{4}\end{aligned}[/tex]
Kesimpulan jawaban :
∴ Hasil rasional dari [tex]\displaystyle\sf\frac{\sqrt{5-2\sqrt6}}{\sqrt{3-\sqrt5}}[/tex] adalah
[tex]\displaystyle=\sf\frac{3\sqrt{9-3\sqrt5}+\sqrt{45-15\sqrt5} -2\sqrt5-2}{4}[/tex]
Pelajari lebih lanjut : Materi Rasional Bentuk akar