Question-145660

Question Number 145660 by imjagoll last updated on 07/Jul/21

Answered by Olaf_Thorendsen last updated on 08/Jul/21

!4 = 4!((1/(0!))−(1/(1!))+(1/(2!))−(1/(3!))+(1/(4!))) !4 = 9 !5 = 5!((1/(0!))−(1/(1!))+(1/(2!))−(1/(3!))+(1/(4!))−(1/(5!))) !5 = 44 !7 = 7!((1/(0!))−(1/(1!))+(1/(2!))−(1/(3!))+(1/(4!))−(1/(5!))+(1/(6!))−(1/(7!))) !7 = 1854 6! = 720 x = (√(!7−6!+!4^2 +(√(!5×3−2^5 )))) x = (√(1854−720+81+(√(132−32)))) x = (√(1225)) = (√(35^2 )) = 35

$$!\mathrm{4}\:=\:\mathrm{4}!\left(\frac{\mathrm{1}}{\mathrm{0}!}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}\right) \\ $$$$!\mathrm{4}\:=\:\mathrm{9} \\ $$$$!\mathrm{5}\:=\:\mathrm{5}!\left(\frac{\mathrm{1}}{\mathrm{0}!}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}−\frac{\mathrm{1}}{\mathrm{5}!}\right) \\ $$$$!\mathrm{5}\:=\:\mathrm{44} \\ $$$$!\mathrm{7}\:=\:\mathrm{7}!\left(\frac{\mathrm{1}}{\mathrm{0}!}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}−\frac{\mathrm{1}}{\mathrm{5}!}+\frac{\mathrm{1}}{\mathrm{6}!}−\frac{\mathrm{1}}{\mathrm{7}!}\right) \\ $$$$!\mathrm{7}\:=\:\mathrm{1854} \\ $$$$\mathrm{6}!\:=\:\mathrm{720} \\ $$$${x}\:=\:\sqrt{!\mathrm{7}−\mathrm{6}!+!\mathrm{4}^{\mathrm{2}} +\sqrt{!\mathrm{5}×\mathrm{3}−\mathrm{2}^{\mathrm{5}} }} \\ $$$${x}\:=\:\sqrt{\mathrm{1854}−\mathrm{720}+\mathrm{81}+\sqrt{\mathrm{132}−\mathrm{32}}} \\ $$$${x}\:=\:\sqrt{\mathrm{1225}}\:=\:\sqrt{\mathrm{35}^{\mathrm{2}} }\:=\:\mathrm{35} \\ $$

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Question-145660

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