1. Sekolah Dasar
  2. Sekolah Menengah Pertama
  3. Sekolah Menengah Atas
  4. Mata Pelajaran
  5. Programming

Quiz! Kayaknya lagi musim materi Merasionalkan Penyebut Mapel MTK ini

Berikut ini adalah pertanyaan dari BNP999 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Quiz!Kayaknya lagi musim materi Merasionalkan Penyebut Mapel MTK ini ƪ(˘⌣˘)ʃ.
Ada soal! Ini kerjakan!
.
Tolong sederhanakan!
\sf \frac{ \sqrt{2} + \sqrt{ 3} }{ \sqrt{5} + \sqrt{6} + \sqrt{7} } \\
​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawab:

=\displaystyle\bf\frac{1}{52}(4\sqrt{10}+6\sqrt{3}+2\sqrt{14}+4\sqrt{15}+9\sqrt{2}+2\sqrt{21}-2\sqrt{105}-3\sqrt{70})

Penjelasan:
\sf\frac{a}{b+\sqrt{c}}=\frac{a(b-\sqrt{c})}{b^2-c}\\\sf\frac{a}{b-\sqrt{c}}=\frac{a(b+\sqrt{c})}{b^2-c}
maka

\frac{\sqrt{2}+\sqrt{3}}{\sqrt{5}+\sqrt{6}+\sqrt{7}}=\frac{(\sqrt{2}+\sqrt{3})(\sqrt{5}-(\sqrt{6}+\sqrt{7}))}{(\sqrt{5}+(\sqrt{6}+\sqrt{7}))(\sqrt{5}-(\sqrt{6}+\sqrt{7}))}\\\\=\frac{\sqrt{2\cdot5}-\sqrt{2\cdot6}-\sqrt{2\cdot7}+\sqrt{3\cdot5}-\sqrt{3\cdot6}-\sqrt{3\cdot7}}{(\sqrt5)^2-(\sqrt{6}+\sqrt{7})^2}\\\\=\frac{\sqrt{10}-\sqrt{12}-\sqrt{14}+\sqrt{15}-\sqrt{18}-\sqrt{21}}{5-((\sqrt{6})^2+2(\sqrt{6})(\sqrt{7})+(\sqrt{7})^2)}

=\frac{\sqrt{10}-\sqrt{4\cdot3}-\sqrt{14}+\sqrt{15}-\sqrt{9\cdot2}-\sqrt{21}}{5-(6+2\sqrt{42}+7)}\\\\=\frac{\sqrt{10}-\sqrt{4}\sqrt{3}-\sqrt{14}+\sqrt{15}-\sqrt{9}\sqrt{2}-\sqrt{21}}{5-6-2\sqrt{42}-7}\\\\=\frac{\sqrt{10}-2\sqrt{3}-\sqrt{14}+\sqrt{15}-3\sqrt{2}-\sqrt{21}}{-8-2\sqrt{42}}\\\\=\frac{\sqrt{10}-2\sqrt{3}-\sqrt{14}+\sqrt{15}-3\sqrt{2}-\sqrt{21}}{-(8+2\sqrt{42})}\\\\=-\frac{\sqrt{10}-2\sqrt{3}-\sqrt{14}+\sqrt{15}-3\sqrt{2}-\sqrt{21}}{8+2\sqrt{42}}

=-\frac{\sqrt{10}-2\sqrt{3}-\sqrt{14}+\sqrt{15}-3\sqrt{2}-\sqrt{21}}{2(4+\sqrt{42})}\\\\=-\frac{(\sqrt{10}-2\sqrt{3}-\sqrt{14}+\sqrt{15}-3\sqrt{2}-\sqrt{21})(4-\sqrt{42})}{2(4+\sqrt{42})(4-\sqrt{42})}\\\\=-\frac{4\sqrt{10}-4(2\sqrt{3})-4\sqrt{14}+4\sqrt{15}-4(3\sqrt{2})-4\sqrt{21}-\sqrt{42\cdot10}+2\sqrt{3\cdot42}+\sqrt{14\cdot42}-\sqrt{15\cdot42}+3\sqrt{2\cdot42}+\sqrt{21\cdot42}}{2(4^2-\sqrt{42}^2)}=-\frac{4\sqrt{10}-8\sqrt{3}-4\sqrt{14}+4\sqrt{15}-12\sqrt{2}-4\sqrt{21}-\sqrt{2\cdot3\cdot7\cdot2\cdot5}+2\sqrt{3\cdot2\cdot3\cdot7}+\sqrt{2\cdot7\cdot2\cdot3\cdot7}-\sqrt{3\cdot5\cdot2\cdot3\cdot7}+3\sqrt{2\cdot2\cdot3\cdot7}+\sqrt{3\cdot7\cdot2\cdot3\cdot7}}{2(16-42)}=-\frac{4\sqrt{10}-8\sqrt{3}-4\sqrt{14}+4\sqrt{15}-12\sqrt{2}-4\sqrt{21}-\sqrt{2^2\cdot3\cdot7\cdot5}+2\sqrt{3^2\cdot2\cdot7}+\sqrt{2^2\cdot7^2\cdot3}-\sqrt{3^2\cdot5\cdot2\cdot7}+3\sqrt{2^2\cdot3\cdot7}+\sqrt{3^2\cdot7^2\cdot2}}{2(-(42-16))}=\frac{4\sqrt{10}-8\sqrt{3}-4\sqrt{14}+4\sqrt{15}-12\sqrt{2}-4\sqrt{21}-\sqrt{2^2}\sqrt{3\cdot7\cdot5}+2\sqrt{3^2}\sqrt{2\cdot7}+\sqrt{2^2}\sqrt{7^2\cdot3}-\sqrt{3^2}\sqrt{5\cdot2\cdot7}+3\sqrt{2^2}\sqrt{3\cdot7}+\sqrt{3^2}\sqrt{7^2}\sqrt{2}}{2(26)}=\frac{4\sqrt{10}-8\sqrt{3}-4\sqrt{14}+4\sqrt{15}-12\sqrt{2}-4\sqrt{21}-2\sqrt{105}+6\sqrt{14}+2(7)\sqrt{3}-3\sqrt{70}+6\sqrt{21}+21\sqrt{2}}{52}

=\frac{4\sqrt{10}-8\sqrt{3}-4\sqrt{14}+4\sqrt{15}-12\sqrt{2}-4\sqrt{21}-2\sqrt{105}+6\sqrt{14}+14\sqrt{3}-3\sqrt{70}+6\sqrt{21}+21\sqrt{2}}{52}

=\frac{4\sqrt{10}+14\sqrt{3}-8\sqrt{3}+6\sqrt{14}-4\sqrt{14}+4\sqrt{15}+21\sqrt{2}-12\sqrt{2}+6\sqrt{21}-4\sqrt{21}-2\sqrt{105}-3\sqrt{70}}{52}
=\displaystyle\bf\frac{1}{52}(4\sqrt{10}+6\sqrt{3}+2\sqrt{14}+4\sqrt{15}+9\sqrt{2}+2\sqrt{21}-2\sqrt{105}-3\sqrt{70})
(xcvi)

Semoga dengan pertanyaan yang sudah terjawab oleh xcvi dapat membantu memudahkan mengerjakan soal, tugas dan PR sekolah kalian.

Apabila terdapat kesalahan dalam mengerjakan soal, silahkan koreksi jawaban dengan mengirimkan email ke yomemimo.com melalui halaman Contact

Last Update: Wed, 09 Nov 22
<< Previous Post
Next Post >>