Question Number 21795 by hi147 last updated on 04/Oct/17 $${help} \\ $$$${x}\in{N} \\ $$$${determine}\:{x}\:{where}\:\mathrm{7}\:{divise}\:\mathrm{2}^{{x}} +\mathrm{3}^{{x}} \\ $$$$ \\ $$ Commented by Rasheed.Sindhi last updated on…
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Question Number 87313 by mr W last updated on 03/Apr/20 $${a},{b},{c}=\mathrm{1},\mathrm{2},\mathrm{3},…,{n} \\ $$$${find}\:\underset{{a}\neq{b}\neq{c}} {\sum}{abc} \\ $$ Commented by mr W last updated on 04/Apr/20 $${yes},\:{you}\:{are}\:{right}.…
Question Number 152844 by bobhans last updated on 02/Sep/21 $$\:\:\:\:\:\:\:\:\:\:\:{simplify}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{3}^{{x}} \:+\:\mathrm{63}}{\mathrm{21}^{{x}−\mathrm{2}} \:+\:\mathrm{7}^{{x}−\mathrm{1}} }\: \\ $$ Answered by Rasheed.Sindhi last updated on 02/Sep/21 $$\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{3}^{{x}}…
Question Number 152840 by liberty last updated on 02/Sep/21 Answered by MJS_new last updated on 04/Sep/21 $$\mathrm{transforming}\:\Rightarrow \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\left({y}^{\mathrm{2}} +\mathrm{3}\right){x}−{y}\left(\mathrm{3}{y}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{1}}{{y}}{x}+{y}^{\mathrm{2}}…
Question Number 87302 by M±th+et£s last updated on 03/Apr/20 $${for}\:\mid{z}−\mathrm{1}\mid=\mathrm{1}\:{show}\:{that} \\ $$$${tan}\left(\frac{{arg}\left({z}−\mathrm{1}\right)}{\mathrm{2}}\right)−\frac{\mathrm{2}{i}}{{z}}=−\mathrm{1} \\ $$ Commented by MJS last updated on 04/Apr/20 $$\mathrm{tan}\:\frac{\mathrm{arg}\:\left({z}−\mathrm{1}\right)}{\mathrm{2}}\:\in\mathbb{R} \\ $$$$−\frac{\mathrm{2i}}{{z}}\notin\mathbb{R}\:\forall\:{z}={a}+{b}\mathrm{i};\:{a}\neq\mathrm{0} \\…
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Question Number 152828 by mathdanisur last updated on 01/Sep/21 $$\underset{\:\mathrm{1}} {\overset{\:\infty} {\int}}\:\frac{\sqrt{\mathrm{x}}\:\mathrm{ln}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:\mathrm{dx}\:=\:? \\ $$ Answered by Ar Brandon last updated on 01/Sep/21 $${f}\left(\kappa\right)=\int_{\mathrm{1}} ^{\infty}…
Question Number 152827 by mathdanisur last updated on 01/Sep/21 Answered by MJS_new last updated on 02/Sep/21 $${p}=\frac{{xy}}{{x}+{y}}\:\Leftrightarrow\:{y}=\frac{{px}}{{x}−{p}}\:\Leftrightarrow\:{x}=\frac{{py}}{{y}−{p}} \\ $$$$\Rightarrow\:{x}>{p}\wedge{y}>{p}\:\wedge\:\left({x}−{p}\right)\mid\left({px}\right)\:\wedge\:\left({y}−{p}\right)\mid\left({py}\right) \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}\right)\:{y}={x}\:\Rightarrow\:{y}={x}=\mathrm{2}{p} \\ $$$$\left(\mathrm{2}\right)\:{y}={p}+\mathrm{1}\:\Rightarrow\:{x}={p}\left({p}+\mathrm{1}\right)…
Question Number 21733 by Joel577 last updated on 02/Oct/17 $$\mathrm{If}\:{p}\:\mathrm{is}\:\mathrm{one}\:\mathrm{of}\:\:\mathrm{roots}\:\mathrm{from}\:{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\:+\:\mathrm{6}\:=\:\mathrm{0} \\ $$$$\mathrm{then}\:{p}^{\mathrm{4}} \:+\:\mathrm{16}{p}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:… \\ $$ Answered by mrW1 last updated on 02/Oct/17 $$\mathrm{p}^{\mathrm{2}} −\mathrm{2p}+\mathrm{6}=\mathrm{0}…