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Question Number 145224 by mathdanisur last updated on 03/Jul/21
(√(2(√3) + 2)) - (√((√3) - (√2))) = ?
$$\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:+\:\mathrm{2}}\:-\:\sqrt{\sqrt{\mathrm{3}}\:-\:\sqrt{\mathrm{2}}}\:=\:? \\ $$
Answered by Olaf_Thorendsen last updated on 03/Jul/21
((√3)+(√2)+1)^2  = 6+2(√6)+2(√3)+2(√2)    x = (√(2(√3)+2))−(√((√3)−(√2)))  x = (1/( (√((√3)+(√2)))))((√((2(√3)+2)((√3)+(√2))))−1)  x = (1/( (√((√3)+(√2)))))((√(6+2(√6)+2(√3)+2(√2)))−1)  x = (1/( (√((√3)+(√2)))))((√(((√3)+(√2)+1)^2 ))−1)  x = (1/( (√((√3)+(√2)))))((√3)+(√2)+1−1)  x = (1/( (√((√3)+(√2)))))((√3)+(√2))  x = (√((√3)+(√2)))
$$\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{6}+\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$ \\ $$$${x}\:=\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}}−\sqrt{\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}} \\ $$$${x}\:=\:\frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}}}\left(\sqrt{\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}\right)\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)}−\mathrm{1}\right) \\ $$$${x}\:=\:\frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}}}\left(\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}}}−\mathrm{1}\right) \\ $$$${x}\:=\:\frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}}}\left(\sqrt{\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{1}\right) \\ $$$${x}\:=\:\frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}}}\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}+\mathrm{1}−\mathrm{1}\right) \\ $$$${x}\:=\:\frac{\mathrm{1}}{\:\sqrt{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}}}\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right) \\ $$$${x}\:=\:\sqrt{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}} \\ $$

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