Question Number 22316 by Tinkutara last updated on 15/Oct/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{p}} \:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{expansion}\:\mathrm{of}\:\left({a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +{a}_{\mathrm{3}} {x}^{\mathrm{3}} +…+{a}_{{k}} {x}^{{k}} \right)^{{n}} \\ $$$$\mathrm{is}\:\Sigma\frac{{n}!}{{n}_{\mathrm{0}} !{n}_{\mathrm{1}} !{n}_{\mathrm{2}}…
Question Number 22315 by Tinkutara last updated on 15/Oct/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{coefficient}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +…+{x}_{{k}} \right)^{{n}} \\ $$$$=\:\frac{{n}!}{\left({q}!\right)^{{k}−{r}} \left[\left({q}+\mathrm{1}\right)!\right]^{{r}} }\:,\:\mathrm{where}\:{n}\:=\:{qk}\:+\:{r}, \\ $$$$\mathrm{0}\:\leqslant\:{r}\:\leqslant\:{k}\:−\:\mathrm{1} \\ $$ Terms…
Question Number 153381 by mathdanisur last updated on 06/Sep/21 $$\mathrm{let}\:\:\mathrm{x};\mathrm{y};\mathrm{z}\geqslant\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{12} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{min}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{S}\:=\:\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:+\:\mathrm{xyz}\:+\:\frac{\mathrm{1}}{\mathrm{xy}\:+\:\mathrm{yz}\:+\:\mathrm{zx}} \\ $$ Commented by mr W last updated…
Question Number 22313 by math solver last updated on 15/Oct/17 Commented by Joel577 last updated on 16/Oct/17 $$\sqrt{\mathrm{5}{x}\:+\:\mathrm{7}}\:−\:\sqrt{\mathrm{3}{x}\:+\:\mathrm{1}}\:=\:\sqrt{{x}\:+\:\mathrm{3}} \\ $$$${x}\:\geqslant\:−\frac{\mathrm{7}}{\mathrm{5}}\:\:\mathrm{and}\:\:{x}\:\geqslant\:−\frac{\mathrm{1}}{\mathrm{3}}\:\:\mathrm{and}\:\:{x}\:\geqslant\:−\mathrm{3} \\ $$$$\Rightarrow\:{x}\:\geqslant\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\…
Question Number 87817 by M±th+et£s last updated on 06/Apr/20 $${f}\left(\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}\right)+{f}\left(\frac{{x}+\mathrm{3}}{{x}−\mathrm{1}}\right)={x} \\ $$$${find}\:{f}\left({x}\right) \\ $$ Commented by M±th+et£s last updated on 06/Apr/20 $${like}\:\mathrm{87755}\:{question}\:{but}\:{here}\:\left({x}−\mathrm{1}\right) \\ $$ Commented…
Question Number 153339 by amin96 last updated on 06/Sep/21 $$\:\:\frac{{x}−\mathrm{1}}{{x}}+\frac{{x}−\mathrm{2}}{{x}}+\frac{{x}−\mathrm{3}}{{x}}+\ldots+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:\:\:\:{x}=? \\ $$$$\because\therefore\because\therefore\because\:\:{Easy}\:{question}\therefore\because\therefore\because\therefore\because \\ $$ Answered by MJS_new last updated on 06/Sep/21 $$\frac{\underset{{j}=\mathrm{1}} {\overset{{x}−\mathrm{1}} {\sum}}{j}}{{x}}=\frac{{x}−\mathrm{1}}{\mathrm{2}}=\mathrm{3}\:\Rightarrow\:{x}=\mathrm{7} \\…
Question Number 22244 by ajfour last updated on 14/Oct/17 $${Solve}\:{the}\:{inequality}\:: \\ $$$$\:−\mathrm{9}\left(\sqrt[{\mathrm{4}}]{{x}}\right)+\sqrt{{x}}+\mathrm{18}\:\geqslant\:\mathrm{0}\:. \\ $$ Answered by $@ty@m last updated on 14/Oct/17 $${Let}\:\sqrt{{x}}={y} \\ $$$$−\mathrm{9}\sqrt{{y}}+{y}+\mathrm{18}\geqslant\mathrm{0} \\…
Question Number 22220 by Tinkutara last updated on 13/Oct/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{integral}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{4log}_{{x}/\mathrm{2}} \left(\sqrt{{x}}\right)+\mathrm{2log}_{\mathrm{4}{x}} \left({x}^{\mathrm{2}} \right)= \\ $$$$\mathrm{3log}_{\mathrm{2}{x}} \left({x}^{\mathrm{3}} \right)\:\mathrm{is} \\ $$ Answered by ajfour last…
Question Number 87755 by john santu last updated on 06/Apr/20 $$\mathrm{f}\left(\frac{\mathrm{x}−\mathrm{3}}{\mathrm{x}+\mathrm{1}}\right)\:+\:\mathrm{f}\left(\frac{\mathrm{x}+\mathrm{3}}{\mathrm{1}−\mathrm{x}}\right)\:=\:\mathrm{x} \\ $$$$\mathrm{find}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$ Commented by john santu last updated on 06/Apr/20 $$\left(\mathrm{i}\right)\:\mathrm{let}\:\frac{\mathrm{x}−\mathrm{3}}{\mathrm{x}+\mathrm{1}}\:=\:\mathrm{x}'\:\Rightarrow\mathrm{f}\left(\mathrm{x}'\right)\:+\:\mathrm{f}\left(\frac{\mathrm{x}'−\mathrm{3}}{\mathrm{x}'+\mathrm{1}}\right)\:=\:\frac{\mathrm{x}'+\mathrm{3}}{\mathrm{1}−\mathrm{x}} \\…
Question Number 22199 by chernoaguero@gmail.com last updated on 13/Oct/17 Commented by chernoaguero@gmail.com last updated on 13/Oct/17 $$\mathrm{Thank}\:\mathrm{you}\:\mathrm{guys} \\ $$ Commented by Rasheed.Sindhi last updated on…