Jawab: [tex]\displaystyle(a.)\:\:\sqrt{2+\sqrt{3}}\:\:\:\:\:\:=\:\frac{\sqrt6+\sqrt2}{2}\\\\(b.)\:\:\sqrt{6+\sqrt{35}}\:\:\:\:=\:\frac{\sqrt{14}+\sqrt{10}}{2}\\\\(c.)\:\:\sqrt{6-\sqrt{27}}\:\:\:\:=\:\frac{3\sqrt{2}-\sqrt{6}}{2}\\\\(d.)\:\:\sqrt{10-\sqrt{21}}\:=\:\frac{\sqrt{2\sqrt{79}+20}-\sqrt{20-2\sqrt{79}}}{2}\\\\(e.)\:\:\frac{2}{\sqrt{2+\sqrt{3}}}\:\:\:\:\:=\:\sqrt{6}-\sqrt{2}\\\\(f.)\:\:\frac{\sqrt2}{\sqrt{6-\sqrt{27}}}\:\:\:=\:\frac{3+\sqrt{3}}{3}[/tex] (Cara selengkapnya terdapat pada gambar)
Penjelasan dengan langkah-langkah: Akar didalam akar, rumusnya:
Jawab:
[tex]\displaystyle(a.)\:\:\sqrt{2+\sqrt{3}}\:\:\:\:\:\:=\:\frac{\sqrt6+\sqrt2}{2}\\\\(b.)\:\:\sqrt{6+\sqrt{35}}\:\:\:\:=\:\frac{\sqrt{14}+\sqrt{10}}{2}\\\\(c.)\:\:\sqrt{6-\sqrt{27}}\:\:\:\:=\:\frac{3\sqrt{2}-\sqrt{6}}{2}\\\\(d.)\:\:\sqrt{10-\sqrt{21}}\:=\:\frac{\sqrt{2\sqrt{79}+20}-\sqrt{20-2\sqrt{79}}}{2}\\\\(e.)\:\:\frac{2}{\sqrt{2+\sqrt{3}}}\:\:\:\:\:=\:\sqrt{6}-\sqrt{2}\\\\(f.)\:\:\frac{\sqrt2}{\sqrt{6-\sqrt{27}}}\:\:\:=\:\frac{3+\sqrt{3}}{3}[/tex]
(Cara selengkapnya terdapat pada gambar)
Penjelasan dengan langkah-langkah:
Akar didalam akar, rumusnya:
[tex]\sqrt{p+q+\sqrt{4pq}}=\sqrt{p}+\sqrt{q}\\\\\sqrt{p+q-\sqrt{4pq}}=\sqrt{p}-\sqrt{q}\:\:\:\{p > q\}[/tex]
Sederhanakan, lalu rasionalkan..
[tex]\displaystyle(a.)\:\:\sqrt{2+\sqrt{3}}\\\\\sqrt{p+q+\sqrt{4pq}}=\sqrt{p}+\sqrt{q}\\\\p+q = 2,\:\:4pq=3\\\\p+q=2,\:\:pq=\frac{3}{4}[/tex]
jika p dan q akar-akar pers. kuadrat x² + Bx + C = 0, maka dengan teori Vieta:
[tex]\displaystyle x^2-(p+q)x =-pq\\\\x^2-2x=-\frac{3}{4}\\\\x^2-2x+\left(\frac{2}{2}\right)^2=-\frac{3}{4}+\left(\frac{2}{2}\right)^2\\\\x^2-2x(1)+1^2=-\frac{3}{4}+1^2\\\\\because\:p^2-2pq+q^2=(p-q)^2\:\therefore\\\\(x-1)^2=-\frac{3}{4}+1^2\\\\(x-1)^2=-\frac{3}{4}+1\\\\(x-1)^2=-\frac{3}{4}+\frac{4}{4}\\\\(x-1)^2=\frac{1}{4}\\\\|x-1|=\sqrt{\frac{1}{4}}\\\\|x-1|=\frac{\sqrt1}{\sqrt4}\\\\|x-1|=\frac{1}{2}\\\\\because|a|=b,\:\:a=\{b, -b\}\therefore[/tex]
[tex]\displaystyle x=\{p,\:\:q\}\\\\x=\left\{\frac{1}{2}+1,\:\:-\frac{1}{2}+1\right\}\\\\x=\left\{\frac{1}{2}+\frac{2}{2},\:\:-\frac{1}{2}+\frac{2}{2}\right\}\\\\x=\left\{\frac{3}{2},\:\:\frac{1}{2}\right\}\\\\p=\frac{3}{2},\:\:q=\frac{1}{2}[/tex]
Oleh karena itu
[tex]\displaystyle\sqrt{p+q+\sqrt{4pq}}=\sqrt{p}+\sqrt{q}\\\\~~~~~~~~~~\sqrt{2+\sqrt{3}}=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\\\\~~~~~~~~~~\sqrt{2+\sqrt{3}}=\frac{\sqrt3}{\sqrt2}+\frac{\sqrt1}{\sqrt2}\\\\~~~~~~~~~~\sqrt{2+\sqrt{3}}=\frac{\sqrt3}{\sqrt2}+\frac{1}{\sqrt2}\\\\~~~~~~~~~~\sqrt{2+\sqrt{3}}=\frac{\sqrt3+1}{\sqrt{2}}\\\\~~~~~~~~~~\sqrt{2+\sqrt{3}}=\frac{(\sqrt3+1)\sqrt{2}}{\sqrt{2}\sqrt{2}}\\\\~~~~~~~~~~\sqrt{2+\sqrt{3}}=\frac{\sqrt3\sqrt{2}+\sqrt{2}}{2}[/tex]
[tex]\displaystyle\sf Maka\\\\~~~~~~~~~~\sqrt{2+\sqrt{3}}=\frac{\sqrt6+\sqrt{2}}{2}[/tex]
Untuk cara dan jawaban pada nomor-nomor lainnya, silakan lihat pada gambar (xcvi)