Question

Bagaimana teknik menyerderhanakan bentuk akar ini?
[tex]\displaystyle \sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}[/tex]

Answer 1

Verified answer

Jawab:
[tex]= \bf\sqrt{10}+\sqrt{2}[/tex]

Penjelasan dengan langkah-langkah:
Misal hasil penyederhanaan = A
[tex]\displaystyle A= \sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\\\\A^2= \left(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\right)^2\\\\\because(a+b)^2=a^2+b^2+2ab\therefore\\\\A^2= \left(\sqrt{8+2\sqrt{10+2\sqrt{5}}}\right)^2+\left(\sqrt{8-2\sqrt{10+2\sqrt{5}}}\right)^2\\\\~~~~~~~~\:+2\left(\sqrt{8+2\sqrt{10+2\sqrt{5}}}\right)\left(\sqrt{8-2\sqrt{10+2\sqrt{5}}}\right)[/tex]

[tex]\displaystyle A^2= 8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}\\\\~~~~~~~~\:+2\sqrt{\left(8+2\sqrt{10+2\sqrt{5}}\right)\left(8-2\sqrt{10+2\sqrt{5}}\right)}\\\\\because(a+b)(a-b)=a^2-b^2\therefore[/tex]

[tex]\displaystyle A^2= 8+8+2\sqrt{10+2\sqrt{5}}-2\sqrt{10+2\sqrt{5}}\\\\~~~~~~~~\:+2\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}\\\\ A^2= 16+2\sqrt{64-2^2\left(\sqrt{10+2\sqrt{5}}\right)^2}\\\\ A^2= 16+2\sqrt{64-4\left(10+2\sqrt{5}\right)}\\\\ A^2= 16+2\sqrt{64-40-8\sqrt{5}\right)}\\\\ A^2= 16+2\sqrt{24-8\sqrt{5}}\\\\A^2= 16+2\sqrt{8(3-\sqrt{5})}\\\\A^2= 16+2\sqrt{8}\sqrt{3-\sqrt{5}}\\\\A^2= 16+2\sqrt{4}\sqrt{2}\sqrt{3-\sqrt{5}}\\\\A^2= 16+2(2)\sqrt{2}\sqrt{3-\sqrt{5}}[/tex]

[tex]\displaystyle A^2= 16+4\sqrt{2}\sqrt{3-\sqrt{5}}\\\\A^2= 16+4\sqrt{2(3-\sqrt{5})}\\\\A^2= 16+4\sqrt{6-2\sqrt{5}}\\\\\because\sqrt{p+q\pm2\sqrt{pq}}=\sqrt{p}\pm\sqrt{q}\:\:\{p > q\}\therefore\\\rightarrow\sqrt{6-2\sqrt{5}}\rightarrow p+q=6,\:\:\:p\:q\:\:=5\\~~~~~~~~~~~~~~~~~~~~~~~5+1=6,\:\:\:5(1)=5\\\rightarrow\sqrt{6-2\sqrt{5}}=\sqrt{5}-\sqrt{1}\\~~~~~~~~~~~~~~~~~~~=\sqrt{5}-1[/tex]

Oleh karena itu

[tex]\displaystyle A^2= 16+4\left(\sqrt{5}-1\right)\\\\A^2= 16+4\sqrt{5}-4\\\\A^2= 12+4\sqrt{5}\\\\A=\sqrt{12+4\sqrt{5}}[/tex]

Hasil =
[tex]\displaystyle\sqrt{12+4\sqrt{5}}[/tex]

Sederhanakan lagi

[tex]\displaystyle\because\sqrt{p+q\pm2\sqrt{pq}}=\sqrt{p}\pm\sqrt{q}\:\:\{p > q\}\therefore\\\rightarrow\sqrt{12+4\sqrt{5}}\\\rightarrow\sqrt{12+2(2)\sqrt{5}}\\\rightarrow\sqrt{12+2\sqrt{4}\sqrt{5}}\\\rightarrow\sqrt{12+2\sqrt{4(5)}}\\\rightarrow\sqrt{12+2\sqrt{20}}\:\:\:\:\rightarrow p+q=12,\:\:\:\:p\:q\:\:=20\\~~~~~~~~~~~~~~~~~~~~~~~~\:~~~10+2=12,\:10(2)=20[/tex]

[tex]\displaystyle \rightarrow\sqrt{12+4\sqrt{5}} = \bf\sqrt{10}+\sqrt{2}[/tex]   ✔️

(xcvi)