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If-x-y-gt-0-then-prove-that-3-x-2-y-2-2-xy-2-x-y-

Question Number 137208 by bemath last updated on 31/Mar/21 $$\mathrm{If}\:\mathrm{x},\mathrm{y}\:>\:\mathrm{0}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\:\mathrm{3}\sqrt{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }{\mathrm{2}}}\:+\:\sqrt{\mathrm{xy}}\:\geqslant\:\mathrm{2}\left(\mathrm{x}+\mathrm{y}\right)\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Show-that-of-all-rectangles-inscribed-in-a-given-circle-the-square-has-a-maximum-area-

Question Number 6136 by sanusihammed last updated on 15/Jun/16 $${Show}\:{that}\:{of}\:{all}\:{rectangles}\:{inscribed}\:{in}\:{a}\:{given}\:{circle}\: \\ $$$${the}\:{square}\:{has}\:{a}\:{maximum}\:{area}. \\ $$ Answered by Rasheed Soomro last updated on 15/Jun/16 $${All}\:{the}\:{rectangles}\:{inscribed}\:{in}\:{same}\:{circle} \\ $$$${have}\:{equal}\:{diagonals}\:{and}\:{vice}\:{versa}.…

Of-all-rectangular-boxes-without-a-lid-and-having-a-given-surface-area-Find-the-one-with-maximum-volume-

Question Number 6135 by sanusihammed last updated on 15/Jun/16 $${Of}\:{all}\:{rectangular}\:{boxes}\:{without}\:{a}\:{lid}\:{and}\:{having}\:{a}\:{given}\: \\ $$$${surface}\:{area}\:.\:{Find}\:{the}\:{one}\:{with}\:{maximum}\:{volume}. \\ $$ Commented by FilupSmith last updated on 15/Jun/16 $$\mathrm{Edge}\:\mathrm{lengths}\:{a},\:{b},\:{c} \\ $$$$\mathrm{Max}\:\mathrm{volume}\:\mathrm{when}\:{a}={b}={c} \\…

Find-all-positive-integers-n-for-which-there-exist-non-negative-integer-a-1-a-2-a-3-a-n-Such-that-1-2-a-1-1-2-a-2-1-2-a-n-1-3-a-1-2-3-a-2-

Question Number 6133 by sanusihammed last updated on 15/Jun/16 $${Find}\:{all}\:{positive}\:{integers}\:{n}\:{for}\:{which}\:{there}\:{exist}\:{non}\:{negative} \\ $$$${integer}\:\:{a}_{\mathrm{1}\:} ,\:{a}_{\mathrm{2}} \:,\:{a}_{\mathrm{3}} \:,\:\:…..\:,\:{a}_{{n}\:} .\:\:\:{Such}\:{that}\:. \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{{a}_{\mathrm{1}} } }+\frac{\mathrm{1}}{\mathrm{2}^{{a}_{\mathrm{2}} } }+…..+\frac{\mathrm{1}}{\mathrm{2}^{{a}_{{n}} } }\:\:=\:\:\frac{\mathrm{1}}{\mathrm{3}^{{a}_{\mathrm{1}} }…

Question-137206

Question Number 137206 by JulioCesar last updated on 31/Mar/21 Answered by bemath last updated on 31/Mar/21 $$\mathrm{by}\:\mathrm{parts}\:\begin{cases}{\mathrm{u}=\mathrm{ln}\:\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}\right) \:\mathrm{du}=\frac{\mathrm{2}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)}\mathrm{dx}}\\{\mathrm{v}\:=\:\mathrm{x}}\end{cases} \\ $$$$\mathrm{I}\:=\:\mathrm{x}\:\mathrm{ln}\:\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}\right)−\int\:\frac{\mathrm{2x}}{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)}\mathrm{dx} \\ $$$$\mathrm{I}=\mathrm{x}\:\mathrm{ln}\:\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}\right)−\left[\int\:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}\mathrm{dx}+\int\:\frac{\mathrm{1}}{\mathrm{x}+\mathrm{1}}\mathrm{dx}\:\right] \\ $$$$\mathrm{I}=\:\mathrm{x}\:\mathrm{ln}\:\left(\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}\right)−\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\:+\:\mathrm{C}…

Evaluate-the-integral-of-x-x-3-2-x-5-2-4-x-7-2-4-6-1-x-2-2-2-x-4-2-2-4-2-x-6-2-2-4-2-6-2-dx-for-0-lt-x-lt-The-answer-is-saying-

Question Number 6132 by sanusihammed last updated on 15/Jun/16 $${Evaluate}\:{the}\:{integral}\:{of}\:… \\ $$$$ \\ $$$$\left[\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{{x}^{\mathrm{5}} }{\mathrm{2}.\mathrm{4}}−\frac{{x}^{\mathrm{7}} }{\mathrm{2}.\mathrm{4}.\mathrm{6}}+….\right)\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }+\frac{{x}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{2}} .\mathrm{4}^{\mathrm{2}} }−\frac{{x}^{\mathrm{6}} }{\mathrm{2}^{\mathrm{2}} .\mathrm{4}^{\mathrm{2}} .\mathrm{6}^{\mathrm{2}}…