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n-15-find-4n-2-4n-120-n-4-2n-3-n-2-




Question Number 213047 by hardmath last updated on 29/Oct/24
n = 15  find:     ((4n^2   +  4n  +  120)/( (√(n^4   +  2n^3   +  n^2 ))))  =  ?
$$\boldsymbol{\mathrm{n}}\:=\:\mathrm{15} \\ $$$$\mathrm{find}:\:\:\:\:\:\frac{\mathrm{4n}^{\mathrm{2}} \:\:+\:\:\mathrm{4n}\:\:+\:\:\mathrm{120}}{\:\sqrt{\mathrm{n}^{\mathrm{4}} \:\:+\:\:\mathrm{2n}^{\mathrm{3}} \:\:+\:\:\mathrm{n}^{\mathrm{2}} }}\:\:=\:\:? \\ $$
Answered by mehdee7396 last updated on 29/Oct/24
(((2n+1)^2 +119)/(n(n+1)))=4.5
$$\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{119}}{{n}\left({n}+\mathrm{1}\right)}=\mathrm{4}.\mathrm{5} \\ $$
Answered by MrGaster last updated on 29/Oct/24
((4∙15^2 +4∙15+120)/( (√(15^4 +2∙15^3 +15^2 ))))  ((4∙225+60+120)/( (√(50625+6750+225))))  ((900+60+120)/( (√(57600))))  ((1080)/(240))  =4.5
$$\frac{\mathrm{4}\centerdot\mathrm{15}^{\mathrm{2}} +\mathrm{4}\centerdot\mathrm{15}+\mathrm{120}}{\:\sqrt{\mathrm{15}^{\mathrm{4}} +\mathrm{2}\centerdot\mathrm{15}^{\mathrm{3}} +\mathrm{15}^{\mathrm{2}} }} \\ $$$$\frac{\mathrm{4}\centerdot\mathrm{225}+\mathrm{60}+\mathrm{120}}{\:\sqrt{\mathrm{50625}+\mathrm{6750}+\mathrm{225}}} \\ $$$$\frac{\mathrm{900}+\mathrm{60}+\mathrm{120}}{\:\sqrt{\mathrm{57600}}} \\ $$$$\frac{\mathrm{1080}}{\mathrm{240}} \\ $$$$=\mathrm{4}.\mathrm{5} \\ $$
Commented by hardmath last updated on 03/Nov/24
thank you dear professors
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors} \\ $$

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