Quiz! Kayaknya lagi musim materi Merasionalkan Penyebut Mapel MTK ini ƪ(˘⌣˘)ʃ.Ada soal! Ini kerjakan.... Question from @BNP999

August 2022 1 46 Report

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

xcvi

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)