Question Number 143431 by enter last updated on 14/Jun/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 143427 by bramlexs22 last updated on 14/Jun/21 $${if}\:{f}\left({x}\right)\:{is}\:{polynomial}\:{satisfying} \\ $$$${f}\left({x}\right){f}\left(\frac{\mathrm{1}}{{x}}\right)−\mathrm{2}{f}\left({x}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{5} \\ $$$${and}\:{f}\left(\mathrm{2}\right)=\mathrm{14}\:{then}\:{f}\left(\mathrm{3}\right)=? \\ $$ Answered by qaz last updated on 14/Jun/21 $$\begin{cases}{\mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)−\mathrm{2f}\left(\mathrm{x}\right)+\mathrm{2f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)=\mathrm{5}…….\left(\mathrm{1}\right)}\\{\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{f}\left(\mathrm{x}\right)−\mathrm{2f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{2f}\left(\mathrm{x}\right)=\mathrm{5}…….\left(\mathrm{2}\right)}\end{cases} \\…
Question Number 77885 by behi83417@gmail.com last updated on 11/Jan/20 $$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\::\:\boldsymbol{\mathrm{x}} \\ $$$$\mathrm{1}.\sqrt{\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}\right)\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{b}}\right)}+\sqrt{\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{b}}\right)\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{c}}\right)}+\sqrt{\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{c}}\right)\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}\right)}=\boldsymbol{\mathrm{d}} \\ $$$$\left[\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}},\boldsymbol{\mathrm{d}}\in\boldsymbol{\mathrm{R}}\right. \\ $$$$\left.\mathrm{try}\:\mathrm{for}:\:\:\mathrm{a}=\mathrm{4},\mathrm{b}=\mathrm{3},\mathrm{c}=\mathrm{2},\mathrm{d}=\mathrm{1}\right] \\ $$$$\mathrm{2}.\:\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}}+\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}\right)\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}^{\mathrm{2}} }=\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{a}}+\mathrm{1} \\ $$$$\mathrm{3}.\:\left(\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{a}}^{\mathrm{2}} \right)\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{a}}}+\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}}…
Question Number 143418 by help last updated on 14/Jun/21 Answered by physicstutes last updated on 14/Jun/21 $$\mathrm{3}.\mathrm{1}\:{y}\:=\:\mathrm{cos}^{\mathrm{2}} {x}^{\mathrm{2}} +\left(\mathrm{3}−\sqrt{{x}}\right)^{\mathrm{30}} −\mathrm{2}^{{x}} \\ $$$$\mathrm{Let}\:{y}_{\mathrm{1}} \:=\:\mathrm{cos}^{\mathrm{2}} {x}^{\mathrm{2}} ,\:\:{y}_{\mathrm{1}}…
Question Number 77883 by TawaTawa last updated on 11/Jan/20 Commented by TawaTawa last updated on 12/Jan/20 $$ \\ $$ Commented by mathmax by abdo last…
Question Number 77881 by TawaTawa last updated on 11/Jan/20 Commented by TawaTawa last updated on 11/Jan/20 $$\mathrm{Please}\:\mathrm{help}.\:\:\mathrm{14},\:\mathrm{15},\:\mathrm{16},\:\mathrm{17} \\ $$ Commented by abdomathmax last updated on…
Question Number 143414 by Snail last updated on 14/Jun/21 $${Evaluate}\: \\ $$$$\lfloor\frac{\left\{\underset{{i}=\mathrm{1}} {\overset{\mathrm{1010}} {\sum}}\:{tan}^{\mathrm{2}} \left(\frac{{i}\pi}{\mathrm{2021}}\right)\right\}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\underset{{i}=\mathrm{1}} {\overset{\mathrm{1010}} {\prod}}\:{tan}\:\left(\frac{{i}\pi}{\mathrm{2021}}\right)}\rfloor\:\:\:\:{wherw}\lfloor\centerdot\rfloor\:{denotes}\:{GIF} \\ $$$$ \\ $$$$ \\ $$$$ \\…
Question Number 143410 by mathdanisur last updated on 14/Jun/21 $${if}\:\:{x};{y};{z}>\mathrm{0}\:\:{prove}\:{that}… \\ $$$$\left(\frac{{x}}{{z}}\right)^{\mathrm{2}} \boldsymbol{{e}}^{\left(\frac{{z}}{{x}}\right)^{\mathrm{2}} } +\:\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} \boldsymbol{{e}}^{\left(\frac{{x}}{{y}}\right)^{\mathrm{2}} } +\:\left(\frac{{z}}{{y}}\right)^{\mathrm{2}} \:\boldsymbol{{e}}^{\left(\frac{{y}}{{z}}\right)^{\mathrm{2}} } \geqslant\:\mathrm{3}\boldsymbol{{e}} \\ $$ Commented by…
Question Number 77864 by Pratah last updated on 11/Jan/20 Commented by Pratah last updated on 11/Jan/20 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{inequality} \\ $$ Answered by key of knowledge last…
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