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)

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Last Update: Wed, 09 Nov 22