Question Number 164787 by cortano1 last updated on 22/Jan/22 Answered by mr W last updated on 22/Jan/22 $${xy}\left[\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}\right]=\frac{\mathrm{10}\left({x}+{y}\right)^{\mathrm{2}} }{\mathrm{9}} \\ $$$${xy}\left[\left({x}+{y}\right)^{\mathrm{3}} −\mathrm{3}{xy}\left({x}+{y}\right)\right]=\frac{\mathrm{2}\left({x}+{y}\right)^{\mathrm{3}} }{\mathrm{3}} \\…
Question Number 164786 by mathlove last updated on 21/Jan/22 Answered by MJS_new last updated on 22/Jan/22 $$\frac{{A}}{{B}}=\frac{{x}+\mathrm{3}}{{x}^{\mathrm{6}} +\mathrm{2}} \\ $$ Terms of Service Privacy Policy…
Question Number 33702 by math khazana by abdo last updated on 22/Apr/18 $${let}\:{p}\left({x}\right)\:={a}_{\mathrm{0}} \:+{a}_{\mathrm{1}} {x}\:+{a}_{\mathrm{2}} {x}^{\mathrm{2}} \:+…{a}_{{n}} {x}^{{n}} \\ $$$${prove}\:{that}\:\:{a}_{{k}} =\:\frac{{p}^{\left({k}\right)} \left(\mathrm{0}\right)}{{k}!}\:\:\forall\:{k}\:\in\left[\left[\mathrm{0},{n}\right]\right]\:. \\ $$ Commented…
Question Number 99218 by 9027201563 last updated on 19/Jun/20 Commented by 9027201563 last updated on 19/Jun/20 $${pls}\:{k}\:{need}\:{solution}\:{immediately} \\ $$ Commented by bobhans last updated on…
Question Number 99175 by Rio Michael last updated on 19/Jun/20 $$\:\:\mathrm{A}\:\mathrm{man}\:\mathrm{and}\:\mathrm{a}\:\mathrm{woman}\:\mathrm{have}\:\mathrm{3}\:\mathrm{boys}\:\mathrm{and}\:\mathrm{3}\:\mathrm{girls}.\: \\ $$$$\left(\mathrm{i}\right)\:\mathrm{in}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{will}\:\mathrm{they}\:\mathrm{sit}\:\mathrm{in}\:\mathrm{a}\:\mathrm{row}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{the}\:\mathrm{3}\:\mathrm{boys}\:\mathrm{and}\:\mathrm{3}\:\mathrm{girls}\:\mathrm{are}\:\mathrm{inbetween}\:\mathrm{the}\:\mathrm{man}\:\mathrm{and}\:\mathrm{woman}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{Say}\:\mathrm{the}\:\mathrm{man}\:\mathrm{decides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{3}\:\mathrm{boys}\:\mathrm{and}\:\mathrm{3}\:\mathrm{girls}\:\mathrm{he}\:\mathrm{has}\:\mathrm{2}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{kids}\:\mathrm{should}\:\mathrm{help}\:\mathrm{him}\:\mathrm{out}\:\mathrm{in}\:\mathrm{a}\:\mathrm{project}, \\ $$$$\mathrm{in}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{done},\:\mathrm{if}\:\mathrm{heyoungest}\:\mathrm{boy}\:\mathrm{and}\:\mathrm{oldest} \\ $$$$\mathrm{girl}\:\mathrm{can}'\mathrm{t}\:\mathrm{join}. \\ $$…
Question Number 99171 by Quvonchbek last updated on 19/Jun/20 Commented by MJS last updated on 19/Jun/20 $${A}={B}\:\Rightarrow\:\frac{{A}}{{B}}=\mathrm{1}\:\Rightarrow\:\left(\frac{{A}}{{B}}\right)^{\mathrm{2020}} =\mathrm{1} \\ $$ Commented by Quvonchbek last updated…
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Question Number 99117 by M±th+et+s last updated on 18/Jun/20 $${let}\:{a},{b},{c}\:\in\mathbb{R}\:{determine}\:{the}\:{minimum} \\ $$$${value} \\ $$$$ \\ $$$$\frac{\mathrm{3}{a}}{{b}+{c}}+\frac{\mathrm{4}{b}}{{a}+{c}}+\frac{\mathrm{5}{c}}{{a}+{b}} \\ $$ Answered by MJS last updated on 19/Jun/20…
Question Number 164650 by cortano1 last updated on 20/Jan/22 $$\:\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{9}}\:−\sqrt[{\mathrm{3}}]{{x}−\mathrm{9}}\:=\:\mathrm{3}\: \\ $$$$\:{x}=? \\ $$ Answered by mr W last updated on 20/Jan/22 $$\left(\sqrt[{\mathrm{3}}]{\mathrm{9}+{x}}+\sqrt[{\mathrm{3}}]{\mathrm{9}−{x}}\right)^{\mathrm{3}} =\mathrm{3}^{\mathrm{3}} \\…
Question Number 33569 by Joel578 last updated on 19/Apr/18 $$\mathrm{Given}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:+\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c} \\ $$$$\mathrm{with}\:{a},\:{b},\:{c}\:\in\:\mathbb{R},\:\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:{x}_{\mathrm{1}} ,\:{x}_{\mathrm{2}} ,\:{x}_{\mathrm{3}} \:\in\:\mathbb{R} \\ $$$$\mathrm{Let}\:\lambda\:\mathrm{is}\:\mathrm{an}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{that}\:\mathrm{satisfied} \\ $$$${x}_{\mathrm{2}} \:−\:{x}_{\mathrm{1}} \:=\:\lambda \\ $$$${x}_{\mathrm{3}}…