Question

Bagaimana cara menyerderhanakan bentuk akar ini?
[tex]\displaystyle \frac{\sqrt{2-\sqrt{3}}}{3-\sqrt{3}}[/tex]

Answer 1

Verified answer

Jawab:
[tex]\displaystyle=\bf\frac{\sqrt{6}}{6}[/tex]

Penjelasan:
Cara menyederhanakan pecahan
[tex]\displaystyle\frac{\sqrt{2-\sqrt{3}}}{3-\sqrt{3}}[/tex]
Fokus ke pembilangnya
[tex]\displaystyle\sqrt{\:\:\:\:\:2\:\:\:-\:\:\:\sqrt{3}}\:\:=\sqrt{p}-\sqrt{q}\\\sqrt{p+q\pm\sqrt{4pq}}=\sqrt{p}\pm\sqrt{q}[/tex]
{p > q}
Oleh karena itu
[tex]\displaystyle p+q=2,\:\:\:4pq=3\\\\~~~~~~~~~~~~~~~~~~pq=\frac{3}{4}[/tex]
Menurut teori vieta, jika p dan q
akar-akar pers. kuadrat dengan
A = 1
[tex]\displaystyle x^2-(p+q)x+pq=0\\\\x^2-2x+\frac{3}{4}=0\\\\x^2+Bx+C=0\\\\B=-2,\:\:\:C=\frac{3}{4}[/tex]

Pakai rumus ABC {jika A = 1}
[tex]\displaystyle x_{1,\:2}=\frac{-B\pm\sqrt{B^2-4C} }{2}\\\\x_{1,\:2}=\frac{-(-2)\pm\sqrt{(-2)^2-4\left(\frac{3}{4}\right)} }{2}\\\\x_{1,\:2}=\frac{2\pm\sqrt{4-3} }{2}\\\\x_{1,\:2}=\frac{2\pm\sqrt{1} }{2}\\\\x_{1,\:2}=\frac{2\pm1 }{2}[/tex]
p > q, maka
[tex]\displaystyle p=\frac{2+1}{2},\:\:q=\frac{2-1}{2}\\\\p=\frac{3}{2},\:\:\:\:\:\:\:\:\:\:q=\frac{1}{2}[/tex]
Oleh karena itu
[tex]\displaystyle\sqrt{\:\:\:\:\:2\:\:\:-\:\:\:\sqrt{3}}\:\:=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\\\\~~~~~~~~~~~~~~~~~~~~~~=\frac{\sqrt3}{\sqrt2}-\frac{\sqrt1}{\sqrt2}\\\\~~~~~~~~~~~~~~~~~~~~~~=\frac{\sqrt3-1}{\sqrt2}\\\\~~~~~~~~~~~~~~~~~~~~~~=\frac{(\sqrt2)(\sqrt3-1)}{(\sqrt2)^2}\\\\~~~~~~~~~~~~~~~~~~~~~~=\frac{\sqrt{2\cdot3}-\sqrt{2}}{2}\\\\~~~~~~~~~~~~~~~~~~~~~~=\frac{\sqrt{6}-\sqrt{2}}{2}[/tex]
Oleh karena itu
[tex]\displaystyle\frac{\sqrt{2-\sqrt{3}}}{3-\sqrt{3}}=\frac{\frac{\sqrt{6}-\sqrt{2}}{2}}{3-\sqrt{3}}=\frac{\sqrt{6}-\sqrt{2}}{2(3-\sqrt{3})}\\\\=\frac{\sqrt{6}-\sqrt{2}}{6-2\sqrt{3}}=\frac{(\sqrt{6}-\sqrt{2})(6+2\sqrt{3})}{6^2-(2\sqrt{3})^2}\\\\=\frac{6\sqrt{6}+2\sqrt{3}\sqrt{6}-6\sqrt{2}-2\sqrt{3}\sqrt{2}}{36-(2^2)(\sqrt{3})^2}\\\\=\frac{6\sqrt{6}+2\sqrt{3}\sqrt{3}\sqrt{2}-6\sqrt{2}-2\sqrt{6}}{36-4(3)}\\\\=\frac{6\sqrt{6}+2(3)\sqrt{2}-6\sqrt{2}-2\sqrt{6}}{36-12}[/tex]

[tex]\displaystyle=\frac{6\sqrt{6}+6\sqrt{2}-6\sqrt{2}-2\sqrt{6}}{24}\\\\=\frac{4\sqrt{6}}{24}[/tex]

[tex]\displaystyle=\bf\frac{\sqrt{6}}{6}[/tex]
(xcvi)