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x-4-5-3-5-48-10-7-4-3-determinant-2x-1-




Question Number 194491 by horsebrand11 last updated on 08/Jul/23
  x=(√(4+(√(5(√3) +5(√(48−10(√(7+4(√3)))))))))       determinant (((2x−1=?)))
$$\:\:\mathrm{x}=\sqrt{\mathrm{4}+\sqrt{\mathrm{5}\sqrt{\mathrm{3}}\:+\mathrm{5}\sqrt{\mathrm{48}−\mathrm{10}\sqrt{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}}}}}\: \\ $$$$\:\:\:\begin{array}{|c|}{\mathrm{2x}−\mathrm{1}=?}\\\hline\end{array} \\ $$
Answered by cortano12 last updated on 08/Jul/23
  x=(√(4+(√(5(√3)+5(√(48−10(√(7+4(√3)))))))))    x=(√(4+(√(5(√3)+5(√(48−10(2+(√3))))))))    x=(√(4+(√(5(√3)+5(√(28−10(√3)))))))    x=(√(4+(√(5(√3)+5(5−(√3))))))    x=(√(4+(√(25)))) = 3     determinant (((2x−1= 5)))
$$\:\:\mathrm{x}=\sqrt{\mathrm{4}+\sqrt{\mathrm{5}\sqrt{\mathrm{3}}+\mathrm{5}\sqrt{\mathrm{48}−\mathrm{10}\sqrt{\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}}}}} \\ $$$$\:\:\mathrm{x}=\sqrt{\mathrm{4}+\sqrt{\mathrm{5}\sqrt{\mathrm{3}}+\mathrm{5}\sqrt{\mathrm{48}−\mathrm{10}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}}} \\ $$$$\:\:\mathrm{x}=\sqrt{\mathrm{4}+\sqrt{\mathrm{5}\sqrt{\mathrm{3}}+\mathrm{5}\sqrt{\mathrm{28}−\mathrm{10}\sqrt{\mathrm{3}}}}} \\ $$$$\:\:\mathrm{x}=\sqrt{\mathrm{4}+\sqrt{\mathrm{5}\sqrt{\mathrm{3}}+\mathrm{5}\left(\mathrm{5}−\sqrt{\mathrm{3}}\right)}} \\ $$$$\:\:\mathrm{x}=\sqrt{\mathrm{4}+\sqrt{\mathrm{25}}}\:=\:\mathrm{3} \\ $$$$\:\:\begin{array}{|c|}{\mathrm{2x}−\mathrm{1}=\:\mathrm{5}}\\\hline\end{array} \\ $$

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