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1-2-3-1-t-2-dt-

Question Number 70074 by cat2315 last updated on 30/Sep/19 $$\int_{\mathrm{1}} ^{\mathrm{2}} \left[\mathrm{3}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right]{dt}= \\ $$ Commented by mathmax by abdo last updated on 30/Sep/19 $${if}\:\left[..\right]{means}\:{the}\:{integr}\:{part}\:\:{we}\:{get}…

Solve-a-e-2x-e-x-1-e-x-e-lt-0-b-4-2-2x-9-2-x-lt-2-c-9-x-4-3-x-1-27-gt-0-

Question Number 70066 by Maclaurin Stickker last updated on 30/Sep/19 $${Solve} \\ $$$$\left.\mathrm{a}\right)\:{e}^{\mathrm{2}{x}} −{e}^{{x}+\mathrm{1}} −{e}^{{x}} +{e}<\mathrm{0} \\ $$$$\left.\mathrm{b}\right)\mathrm{4}.\mathrm{2}^{\mathrm{2}{x}} −\mathrm{9}.\mathrm{2}^{{x}} <−\mathrm{2} \\ $$$$\left.{c}\right)\mathrm{9}^{{x}} −\mathrm{4}.\mathrm{3}^{{x}+\mathrm{1}} +\mathrm{27}>\mathrm{0} \\…

2016-2007-2018-2009-2020-2011-2022-

Question Number 70044 by naka3546 last updated on 30/Sep/19 $$\sqrt{\mathrm{2016}\:+\:\mathrm{2007}\sqrt{\mathrm{2018}\:+\:\mathrm{2009}\sqrt{\mathrm{2020}\:+\:\mathrm{2011}\sqrt{\mathrm{2022}\:+\:\ldots}}}}\:\:=\:\:… \\ $$ Commented by Prithwish sen last updated on 30/Sep/19 $$\left(\mathrm{a}−\mathrm{6}\right)=\sqrt{\left(\mathrm{a}−\mathrm{6}\right)^{\mathrm{2}} }=\sqrt{\mathrm{a}+\left(\mathrm{a}^{\mathrm{2}} −\mathrm{13a}+\mathrm{36}\right)} \\ $$$$\:\:=\:\sqrt{\mathrm{a}+\left(\mathrm{a}−\mathrm{9}\right)\left(\mathrm{a}−\mathrm{4}\right)}\:=\:\sqrt{\mathrm{a}+\left(\mathrm{a}−\mathrm{9}\right)\sqrt{\left(\mathrm{a}−\mathrm{4}\right)^{\mathrm{2}}…

12345689654321233-5333334096-12345679-12345689654321233-5333334096-12345679-1-3-

Question Number 70021 by naka3546 last updated on 30/Sep/19 $$\sqrt[{\mathrm{3}}]{\sqrt{\sqrt{\mathrm{12345689654321233}\:+\:\mathrm{5333334096}\:\sqrt{\mathrm{12345679}}}}\:−\:\sqrt{\sqrt{\mathrm{12345689654321233}\:−\:\mathrm{5333334096}\:\sqrt{\mathrm{12345679}}}}}\:\:=\:\:… \\ $$ Commented by MJS last updated on 30/Sep/19 $$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve} \\ $$$$\sqrt{{a}+{b}\sqrt{{c}}}={p}+{q}\sqrt{{c}}\:\Leftrightarrow\:{a}={p}^{\mathrm{2}} +{cq}^{\mathrm{2}} \wedge{b}=\mathrm{2}{pq} \\…