Question Number 22379 by Tinkutara last updated on 16/Oct/17 $$\mathrm{For}\:\mathrm{each}\:\mathrm{positive}\:\mathrm{integer}\:{n},\:\mathrm{define}\:{a}_{{n}} \:= \\ $$$$\mathrm{20}\:+\:{n}^{\mathrm{2}} ,\:\mathrm{and}\:{d}_{{n}} \:=\:{gcd}\left({a}_{{n}} ,\:{a}_{{n}+\mathrm{1}} \right).\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{all}\:\mathrm{values}\:\mathrm{that}\:\mathrm{are}\:\mathrm{taken}\:\mathrm{by} \\ $$$${d}_{{n}} \:\mathrm{and}\:\mathrm{show}\:\mathrm{by}\:\mathrm{examples}\:\mathrm{that}\:\mathrm{each}\:\mathrm{of} \\ $$$$\mathrm{these}\:\mathrm{values}\:\mathrm{are}\:\mathrm{attained}. \\…
Question Number 153450 by liberty last updated on 07/Sep/21 $$\begin{cases}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} ={x}+{y}+\mathrm{2}}\\{\left({y}+\mathrm{1}\right)^{\mathrm{2}} ={y}+{z}+\mathrm{2}}\\{\left({z}+\mathrm{1}\right)^{\mathrm{2}} ={z}+{x}+\mathrm{2}}\end{cases} \\ $$ Answered by EDWIN88 last updated on 07/Sep/21 Commented by Tawa11…
Question Number 153446 by mathdanisur last updated on 07/Sep/21 $$\mathrm{L}^{−\mathrm{1}} \left\{\frac{\mathrm{s}}{\mathrm{s}^{\mathrm{2}} \:-\:\mathrm{12s}\:+\:\mathrm{40}}\right\}\:=\:? \\ $$ Commented by alisiao last updated on 07/Sep/21 $$=\:{L}^{−\mathrm{1}} \:\left\{\frac{\left({s}−\mathrm{6}\right)}{\left({s}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{4}}\:+\:\frac{\mathrm{6}}{\left({s}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{4}}\right\}…
Question Number 87886 by TawaTawa1 last updated on 06/Apr/20 Commented by jagoll last updated on 07/Apr/20 $$\mathrm{x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{x}\:=\:\mathrm{2}\:\pm\:\sqrt{\mathrm{3}}\: \\ $$$$\mathrm{y}^{\mathrm{2}}…
Question Number 153412 by mathdanisur last updated on 07/Sep/21 $$\mathrm{If}\:\:\:\mathrm{a};\mathrm{b};\mathrm{c};\mathrm{d}\in\left(\mathrm{0};\infty\right)\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{a}^{\mathrm{3}} }{\mathrm{bc}}\:+\:\frac{\mathrm{b}^{\mathrm{3}} }{\mathrm{cd}}\:+\:\frac{\mathrm{c}^{\mathrm{3}} }{\mathrm{da}}\:+\:\frac{\mathrm{d}^{\mathrm{3}} }{\mathrm{ab}}\:\geqslant\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:+\:\mathrm{d} \\ $$ Answered by puissant last updated on 07/Sep/21…
Question Number 153407 by liberty last updated on 07/Sep/21 Answered by Rasheed.Sindhi last updated on 08/Sep/21 $${x}\equiv\mathrm{2}\:\left({mod}\:\mathrm{5}\right)……..\left({i}\right) \\ $$$${x}\equiv\mathrm{1}\:\left({mod}\:\mathrm{3}\right)……..\left({ii}\right) \\ $$$${x}\equiv\mathrm{6}\:\left({mod}\:\mathrm{14}\right)……\left({iii}\right) \\ $$$$\underset{\underset{} {−}} {{x}\equiv\mathrm{5}\underline{\:\left({mod}\:\mathrm{11}\right)…….\left({iv}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\:\:\:\:\:\:\:\:\:}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…
Question Number 22316 by Tinkutara last updated on 15/Oct/17 $$\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 $$\sqrt{\mathrm{5}{x}\:+\:\mathrm{7}}\:−\:\sqrt{\mathrm{3}{x}\:+\:\mathrm{1}}\:=\:\sqrt{{x}\:+\:\mathrm{3}} \\ $$$${x}\:\geqslant\:−\frac{\mathrm{7}}{\mathrm{5}}\:\:\mathrm{and}\:\:{x}\:\geqslant\:−\frac{\mathrm{1}}{\mathrm{3}}\:\:\mathrm{and}\:\:{x}\:\geqslant\:−\mathrm{3} \\ $$$$\Rightarrow\:{x}\:\geqslant\:−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\…