Question Number 172631 by Mikenice last updated on 29/Jun/22 Commented by mr W last updated on 29/Jun/22 $${x}!=\left({x}+\mathrm{1}\right){x}\left({x}−\mathrm{1}\right) \\ $$$${x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)!=\left({x}+\mathrm{1}\right){x}\left({x}−\mathrm{1}\right) \\ $$$$\left({x}−\mathrm{2}\right)!={x}+\mathrm{1} \\ $$$${let}\:{u}={x}−\mathrm{2} \\…
Question Number 172609 by Mikenice last updated on 29/Jun/22 Answered by Rasheed.Sindhi last updated on 29/Jun/22 $$\mathrm{4}^{{x}} +\mathrm{4}^{{x}} =\mathrm{4}^{\mathrm{8}} ……………\left({i}\right) \\ $$$${x}+{y}=\mathrm{20}……………..\left({ii}\right) \\ $$$${y}+{y}=? \\…
Question Number 172611 by Mikenice last updated on 29/Jun/22 Answered by MJS_new last updated on 29/Jun/22 $$\mathrm{I}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{but}\:\mathrm{I}\:\mathrm{guess}\:\mathrm{ir}'\mathrm{s}\:\mathrm{meant}\:\mathrm{to}\:\mathrm{be} \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{4}{x}}{\mathrm{3}}+\frac{\mathrm{84}}{{x}^{\mathrm{2}} +\mathrm{4}{x}}=\mathrm{11} \\ $$$${t}={x}^{\mathrm{2}} +\mathrm{4}{x} \\…
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Question Number 172599 by Mikenice last updated on 29/Jun/22 Answered by Rasheed.Sindhi last updated on 29/Jun/22 $$\underset{\left({i}\right)} {\underbrace{\mathrm{8}{y}^{{x}} −{y}^{\mathrm{2}{x}} =\mathrm{16}}}\:,\:\underset{\left({ii}\right)} {\underbrace{\mathrm{2}^{{x}} ={y}^{\mathrm{2}} }}\:;\:\:\:\:\:{x}=?,{y}=? \\ $$$$\left({i}\right)\Rightarrow\mathrm{8}\left({y}^{\mathrm{2}}…
Question Number 41522 by Tawa1 last updated on 09/Aug/18 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{root}\:\mathrm{of}\:\:\:\mathrm{26}\:−\:\mathrm{15}\sqrt{\mathrm{3}} \\ $$ Commented by math khazana by abdo last updated on 10/Aug/18 $${inside}\:{C}\:\:{z}^{\mathrm{3}} =\mathrm{26}−\mathrm{15}\sqrt{\mathrm{3}}\:\:{let}\:{z}={x}+{iy}\:\Rightarrow \\…
Question Number 41519 by maxmathsup by imad last updated on 08/Aug/18 $${let}\:{z}=\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\:−{i}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$${calculate}\:\mid{z}^{{n}} \mid\:\:{and}\:{arg}\left({z}^{{n}} \right) \\ $$ Answered by alex041103 last updated on 09/Aug/18…
Question Number 41520 by maxmathsup by imad last updated on 08/Aug/18 $${let}\:{Z}\:=\:{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\:+{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)\:{and}\:\:{A}=\:{Z}+{Z}^{\mathrm{2}} \:+{Z}^{\mathrm{4}} \\ $$$${B}={Z}^{\mathrm{3}} \:+{Z}^{\mathrm{5}} \:+{Z}^{\mathrm{6}} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\overset{−} {{A}}={B} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:{A}+{B}\:=−\mathrm{1}\:{and}\:{A}.{B}\:=\mathrm{2} \\ $$$$\left.\mathrm{3}\right)\:{find}\:\:{A}\:{and}\:{B}. \\…
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