Question Number 162785 by john_santu last updated on 01/Jan/22 Answered by Rasheed.Sindhi last updated on 02/Jan/22 $$ \\ $$$$\frac{{x}^{\mathrm{4}} +\mathrm{2}{x}+\mathrm{5}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)}=\left({x}−\mathrm{1}\right)+\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}}{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)} \\ $$$$\frac{{x}^{\mathrm{2}}…
Question Number 162787 by amin96 last updated on 01/Jan/22 $$\boldsymbol{\mathrm{happy}}\:\boldsymbol{\mathrm{new}}\:\boldsymbol{\mathrm{year}} \\ $$$$\left\{\boldsymbol{{a}};\boldsymbol{{b}};\boldsymbol{{c}}\right\}\in\mathbb{Z}−\left\{\mathrm{0}\right\} \\ $$$$\boldsymbol{{p}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{bx}}+\boldsymbol{{c}}\:\:\:\: \\ $$$$\boldsymbol{{p}}\left(\boldsymbol{{a}}\right)=\mathrm{0} \\ $$$$\boldsymbol{{p}}\left(\boldsymbol{{b}}\right)=\mathrm{0} \\ $$$$\boldsymbol{{p}}\left(\mathrm{1}\right)=? \\ $$ Answered by…
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Question Number 97227 by bemath last updated on 07/Jun/20 Commented by bemath last updated on 07/Jun/20 $$\mathrm{thank}\:\mathrm{you}\:\mathrm{both}\: \\ $$ Answered by som(math1967) last updated on…
Question Number 31677 by gyugfeet last updated on 12/Mar/18 $$ \\ $$$$\mathrm{24}{x}^{\mathrm{3}} −\mathrm{26}{x}^{\mathrm{2}} +\mathrm{9}{x}−\mathrm{1}=\mathrm{0}\left({solve}\right) \\ $$ Answered by Joel578 last updated on 12/Mar/18 $$\mathrm{Trial}\:\mathrm{and}\:\mathrm{error} \\…
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Question Number 97206 by O Predador last updated on 07/Jun/20 Answered by john santu last updated on 07/Jun/20 $$\mathrm{let}\:\sqrt{\mathrm{x}+\mathrm{3}}\:=\:\mathrm{t}\:\Rightarrow\mathrm{x}=\mathrm{t}^{\mathrm{2}} −\mathrm{3} \\ $$$$\mathrm{P}\left(\mathrm{t}\right)\:=\:\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{6}}{\mathrm{t}^{\mathrm{2}} −\mathrm{3}}.\Rightarrow\mathrm{P}\left(\mathrm{Q}\left(\mathrm{x}\right)\right)=\:\frac{\mathrm{Q}\left(\mathrm{x}\right)^{\mathrm{2}} −\mathrm{6}}{\mathrm{Q}\left(\mathrm{x}\right)^{\mathrm{2}}…
Question Number 31670 by gunawan last updated on 12/Mar/18 $$\mathrm{how}\:\mathrm{many}\:\mathrm{roots}\:\mathrm{from}\:\mathrm{equation} \\ $$$${ae}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${from}\:{a}>\mathrm{0}\:? \\ $$ Answered by mrW2 last updated on 12/Mar/18…
Question Number 162721 by mnjuly1970 last updated on 31/Dec/21 $$ \\ $$$$\:\:\:\:\:\:\:{calculate}\: \\ $$$$\:\:\:\:\:\:{f}\:\left({x}\:\right)=\:\frac{\:\mathrm{1}}{\mathrm{4}\left(\mathrm{1}+{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)\right)\:}\:+\frac{\mathrm{1}}{\mathrm{9}\left(\mathrm{1}−{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)\right)}\:\:\left(\:{x}\:\neq\:\mathrm{2}{k}\:\pi\:,\:{k}\:\in\:\mathbb{Z}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{f}_{\:{min}} =\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathscr{A}{dapted}\:\mathscr{F}{rom}\:\mathscr{I}{nstagram}\: \\ $$$$ \\ $$ Answered by…