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Express-15-35-21-5-3-2-5-7-in-the-form-a-3-b-7-

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Question Number 148525 by nadovic last updated on 28/Jul/21
Express (((√(15))+(√(35))+(√(21))+5)/( (√3)+2(√5)+(√7))) in the form a(√(3 ))+ b(√7).
$$\:\mathrm{Express}\:\:\frac{\sqrt{\mathrm{15}}+\sqrt{\mathrm{35}}+\sqrt{\mathrm{21}}+\mathrm{5}}{\:\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{5}}+\sqrt{\mathrm{7}}}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form} \\ $$$$\:{a}\sqrt{\mathrm{3}\:}+\:{b}\sqrt{\mathrm{7}}. \\ $$
Answered by Ar Brandon last updated on 28/Jul/21
(((√(15))+(√(35))+(√(21))+5)/( (√3)+2(√5)+(√7))) =(((√3)∙(√5)+(√7)∙(√5)+(√3)∙(√7)+(√5) ^2 )/( (√3)+2(√5)+(√7))) =(((√5) ^2 +((√3)+(√7))(√5)+(√7)∙(√5))/( (√3)+2(√5)+(√7))) =((((√5)+(√3))((√5)+(√7)))/(((√5)+(√3))+((√5)+(√7))))×((((√5)+(√3))−((√5)+(√7)))/(((√5)+(√3))−((√5)+(√7))))
$$\frac{\sqrt{\mathrm{15}}+\sqrt{\mathrm{35}}+\sqrt{\mathrm{21}}+\mathrm{5}}{\:\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{5}}+\sqrt{\mathrm{7}}} \\ $$$$=\frac{\sqrt{\mathrm{3}}\centerdot\sqrt{\mathrm{5}}+\sqrt{\mathrm{7}}\centerdot\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}\centerdot\sqrt{\mathrm{7}}+\sqrt{\mathrm{5}}\overset{\mathrm{2}} {\:}}{\:\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{5}}+\sqrt{\mathrm{7}}} \\ $$$$=\frac{\sqrt{\mathrm{5}}\overset{\mathrm{2}} {\:}+\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{7}}\right)\sqrt{\mathrm{5}}+\sqrt{\mathrm{7}}\centerdot\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{5}}+\sqrt{\mathrm{7}}} \\ $$$$=\frac{\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{7}}\right)}{\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}\right)+\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{7}}\right)}×\frac{\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}\right)−\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{7}}\right)}{\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}\right)−\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{7}}\right)} \\ $$
Answered by MJS_new last updated on 29/Jul/21
((5+(√7)(√5)+(√7)(√3)+(√5)(√3))/( (√7)+2(√5)+(√3)))= =(((5+(√7)(√5)+(√7)(√3)+(√5)(√3))(−(√7)+2(√5)+(√3)))/(((√7)+2(√5)+(√3))(−(√7)+2(√5)+(√3))))= =((2(√7)(√5)(√3)+8(√7)+6(√5)+8(√3))/(16+4(√5)(√3)))= =(((√7)(√5)(√3)+4(√7)+3(√5)+4(√3))/(2(4+(√5)(√3))))= =((((√7)(√5)(√3)+4(√7)+3(√5)+4(√3))(4−(√5)(√3)))/(2(4+(√5)(√3))(4−(√5)(√3))))= =(((√7)+(√3))/2) ⇒ a=b=(1/2)
$$\frac{\mathrm{5}+\sqrt{\mathrm{7}}\sqrt{\mathrm{5}}+\sqrt{\mathrm{7}}\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{7}}+\mathrm{2}\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}}= \\ $$$$=\frac{\left(\mathrm{5}+\sqrt{\mathrm{7}}\sqrt{\mathrm{5}}+\sqrt{\mathrm{7}}\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\sqrt{\mathrm{3}}\right)\left(−\sqrt{\mathrm{7}}+\mathrm{2}\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}\right)}{\left(\sqrt{\mathrm{7}}+\mathrm{2}\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}\right)\left(−\sqrt{\mathrm{7}}+\mathrm{2}\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}\right)}= \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{7}}\sqrt{\mathrm{5}}\sqrt{\mathrm{3}}+\mathrm{8}\sqrt{\mathrm{7}}+\mathrm{6}\sqrt{\mathrm{5}}+\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{16}+\mathrm{4}\sqrt{\mathrm{5}}\sqrt{\mathrm{3}}}= \\ $$$$=\frac{\sqrt{\mathrm{7}}\sqrt{\mathrm{5}}\sqrt{\mathrm{3}}+\mathrm{4}\sqrt{\mathrm{7}}+\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{2}\left(\mathrm{4}+\sqrt{\mathrm{5}}\sqrt{\mathrm{3}}\right)}= \\ $$$$=\frac{\left(\sqrt{\mathrm{7}}\sqrt{\mathrm{5}}\sqrt{\mathrm{3}}+\mathrm{4}\sqrt{\mathrm{7}}+\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{4}\sqrt{\mathrm{3}}\right)\left(\mathrm{4}−\sqrt{\mathrm{5}}\sqrt{\mathrm{3}}\right)}{\mathrm{2}\left(\mathrm{4}+\sqrt{\mathrm{5}}\sqrt{\mathrm{3}}\right)\left(\mathrm{4}−\sqrt{\mathrm{5}}\sqrt{\mathrm{3}}\right)}= \\ $$$$=\frac{\sqrt{\mathrm{7}}+\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{a}={b}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by nadovic last updated on 30/Jul/21
Thank you Sir
$${Thank}\:{you}\:{Sir} \\ $$

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