Question-173933

Question Number 173933 by mathlove last updated on 21/Jul/22

Commented by cortano1 last updated on 21/Jul/22

$$=\mathrm{1} \\ $$

Answered by som(math1967) last updated on 21/Jul/22

(((2+(√3))/2)/(1+(√((4+2(√3))/4)))) +(((2−(√3))/2)/(1−(√((4−2(√3))/4)))) =(((2+(√3))/2)/((2+(√(((√3)+1)^2 )))/2)) +(((2−(√3))/2)/((2−(√(((√3)−1)^2 )))/2)) =((2+(√3))/(3+(√3))) +((2−(√3))/(3−(√3))) =(((2+(√3))(3−(√3))+(2−(√3))(3+(√3)))/(9−3)) =((6−2(√3)+3(√3)−3+6+2(√3)−3(√3)−3)/6) =((12−6)/6)=1

$$\frac{\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{1}+\sqrt{\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}}}}\:+\frac{\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{1}−\sqrt{\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}}}} \\ $$$$=\frac{\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}}}{\frac{\mathrm{2}+\sqrt{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} }}{\mathrm{2}}}\:+\frac{\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}}}{\frac{\mathrm{2}−\sqrt{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} }}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{3}+\sqrt{\mathrm{3}}}\:+\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{3}−\sqrt{\mathrm{3}}} \\ $$$$=\frac{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{3}−\sqrt{\mathrm{3}}\right)+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left(\mathrm{3}+\sqrt{\mathrm{3}}\right)}{\mathrm{9}−\mathrm{3}} \\ $$$$=\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}+\mathrm{6}+\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}}{\mathrm{6}} \\ $$$$=\frac{\mathrm{12}−\mathrm{6}}{\mathrm{6}}=\mathrm{1} \\ $$

Commented by mathlove last updated on 22/Jul/22

$${thanks} \\ $$

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