Question Number 7165 by Tawakalitu. last updated on 15/Aug/16 $$\mathrm{30}\:{teams}\:{participated}\:{in}\:{the}\:{football}\:{tournament}\:.\:{At}\:{the}\: \\ $$$${end}\:{of}\:{the}\:{competition}\:{it}\:{turned}\:{out}\:{that}\:{in} \\ $$$${Any}\:{group}\:{of}\:{three}\:\:{teams}\:{it}\:{is}\:{possible}\:{to}\:{single}\:{out}\:{two} \\ $$$${teams}\:{which}\:{score}\:{equal}\:{point}\:{in}\:{three}\:{games}.\: \\ $$$${within}\:{this}\:{group}\:\left(\mathrm{3}\:{points}\:{are}\:{given}\:{for}\:{the}\:{victory},\:\right. \\ $$$$\left.\mathrm{1}\:{point}\:{for}\:{the}\:{draw}\:,\:\mathrm{0}\:{point}\:{for}\:{the}\:{defeat}\:\right).\: \\ $$$${what}\:{is}\:{the}\:{least}\:{possible}\:{number}\:{of}\:{draws}\:{that}\:{can}\:{occur} \\ $$$${in}\:{such}\:{a}\:{tournament}\:? \\…
Question Number 7161 by Tawakalitu. last updated on 14/Aug/16 $${If}\:\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0},\:\:{prove}\:{that}. \\ $$$${x}\:=\:\frac{\mathrm{2}{c}}{−\:{b}\:\pm\:\sqrt{{b}^{\mathrm{2}} \:−\:\mathrm{4}{ac}}}\: \\ $$ Commented by sou1618 last updated on 14/Aug/16 $$\left(\ast\right)\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}…
Question Number 138235 by mr W last updated on 11/Apr/21 $${for}\:{p},{q}\in\mathbb{R}\:{satisfying}\:{p}^{\mathrm{4}} +{q}^{\mathrm{4}} =\mathrm{4}{pq} \\ $$$${find}\:{the}\:{range}\:{of}\:{p}+{q}\:{when} \\ $$$$\left.\mathrm{1}\right)\:{no}\:{restriction} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{0}\leqslant{p}\leqslant\mathrm{1},\:\mathrm{0}\leqslant{q}\leqslant\mathrm{1} \\ $$ Answered by mr W…
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Question Number 138224 by Bekzod Jumayev last updated on 11/Apr/21 Commented by Bekzod Jumayev last updated on 11/Apr/21 $${Help} \\ $$ Answered by MJS_new last…
Question Number 72689 by aliesam last updated on 31/Oct/19 Answered by Tanmay chaudhury last updated on 01/Nov/19 $${IMO}\:\mathrm{2001}\:{question}… \\ $$$${this}\:{question}\:{appeared}\:{in}\:{IMO}\:\mathrm{2001}…{so}\:{pls}\:{search} \\ $$$${you}\:{will}\:{get}\:{solution} \\ $$ Commented…
Question Number 7147 by Tawakalitu. last updated on 13/Aug/16 Commented by Yozzii last updated on 13/Aug/16 $${u}\left({n}\right)=\left(−\mathrm{1}+\mathrm{2}\right)+\left(−\mathrm{3}+\mathrm{4}\right)+\left(−\mathrm{5}+\mathrm{6}\right)+…+\left(−\left(\mathrm{2}{n}−\mathrm{1}\right)+\mathrm{2}{n}\right) \\ $$$${u}\left({n}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{2}{n}+\mathrm{1}+\mathrm{2}{n}\right)={n} \\ $$$${u}\left(\frac{\mathrm{10}^{\mathrm{9}} }{\mathrm{2}}\right)=\frac{\mathrm{10}^{\mathrm{9}} }{\mathrm{2}}…
Question Number 7138 by Tawakalitu. last updated on 12/Aug/16 Answered by Yozzii last updated on 13/Aug/16 $${x}^{{x}^{\mathrm{2}} } =\mathrm{3}\:\:\left({x}>\mathrm{0}\right) \\ $$$${Let}\:{x}^{\mathrm{2}} ={e}^{{k}} \Rightarrow{x}={e}^{{k}/\mathrm{2}} \\ $$$$\therefore{e}^{\left({k}/\mathrm{2}\right){e}^{{k}}…
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Question Number 7127 by Tawakalitu. last updated on 11/Aug/16 Commented by Yozzii last updated on 11/Aug/16 $${Let}\:{u}={x}\sqrt{\mathrm{1}+\left({x}+\mathrm{1}\right)+\sqrt{\mathrm{1}+\left({x}+\mathrm{2}\right)+\sqrt{\mathrm{1}+\left({x}+\mathrm{3}\right)+\sqrt{\mathrm{1}+….}}}} \\ $$$$\Rightarrow\frac{{u}}{{x}}=\sqrt{\mathrm{1}+\left({x}+\mathrm{1}\right)+\sqrt{\mathrm{1}+\left({x}+\mathrm{2}\right)+\sqrt{\mathrm{1}+\left({x}+\mathrm{3}\right)+\sqrt{\mathrm{1}+….}}}} \\ $$$${let}\:{n}={x}+\mathrm{1}\Rightarrow{x}={n}−\mathrm{1}. \\ $$$$\therefore\frac{{u}}{{n}−\mathrm{1}}=\sqrt{\mathrm{1}+{n}+\sqrt{\mathrm{1}+\left({n}+\mathrm{1}\right)+\sqrt{\mathrm{1}+\left({n}+\mathrm{2}\right)+\sqrt{\mathrm{1}+\left({n}+\mathrm{3}\right)+\sqrt{\mathrm{1}+…..}}}}} \\ $$$$\frac{{u}^{\mathrm{2}}…