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Question Number 197185 by MathematicalUser2357 last updated on 10/Sep/23
Simplify (((((√2))^(√3) ∙((√3))^(√2) +((√2))^(√(12)) )/(((√6))^(√2) +((√2))^((√3)+(√2)) )))^(1/((√3)−(√2)))
$$\mathrm{Simplify} \\ $$$$\sqrt[{\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}}]{\frac{\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{3}}} \centerdot\left(\sqrt{\mathrm{3}}\right)^{\sqrt{\mathrm{2}}} +\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{12}}} }{\left(\sqrt{\mathrm{6}}\right)^{\sqrt{\mathrm{2}}} +\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}} }} \\ $$
Answered by som(math1967) last updated on 10/Sep/23
=^((√3)−(√2)) (√((((√2))^(√3) .((√3))^(√2) +((√2))^(2(√3)) )/(((√3))^(√2) ((√2))^(√2) +((√2))^((√3)+(√2)) ))) =^((√3)−(√2)) (√((((√2))^(√3) {((√3))^(√2) +((√2))^(√3) })/(((√2))^(√2) {((√3))^(√2) +((√2))^(√3) }))) =^((√3)−(√2)) (√(((√2))^((√3)−(√2)) ))=(√2)
$$=\:^{\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}} \sqrt{\frac{\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{3}}} .\left(\sqrt{\mathrm{3}}\right)^{\sqrt{\mathrm{2}}} +\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}\sqrt{\mathrm{3}}} }{\left(\sqrt{\mathrm{3}}\right)^{\sqrt{\mathrm{2}}} \left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{2}}} +\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}} }} \\ $$$$=\:^{\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}} \sqrt{\frac{\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{3}}} \left\{\left(\sqrt{\mathrm{3}}\right)^{\sqrt{\mathrm{2}}} +\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{3}}} \right\}}{\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{2}}} \left\{\left(\sqrt{\mathrm{3}}\right)^{\sqrt{\mathrm{2}}} +\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{3}}} \right\}}} \\ $$$$=\:^{\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}} \sqrt{\left(\sqrt{\mathrm{2}}\right)^{\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}} }=\sqrt{\mathrm{2}} \\ $$

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