Question Number 190731 by 073 last updated on 10/Apr/23 $$\left(\mathrm{4a}^{\mathrm{2}} −\mathrm{19a}−\mathrm{5}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} \mathrm{x}+\mathrm{a}+\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} \mathrm{are}\:\mathrm{roots} \\ $$$$\mathrm{when}\:,\:\mathrm{x}_{\mathrm{1}} <\mathrm{0}\:\:\:,\mathrm{x}_{\mathrm{2}} >\mathrm{0}\:\:,\:\mid\mathrm{x}_{\mathrm{1}} \mid−\mathrm{x}_{\mathrm{2}} >\mathrm{0} \\ $$$$\mathrm{interval}\:\mathrm{of}\:\:\:\mathrm{max}\left(\mathrm{a}\right)=?…
Question Number 190717 by sciencestudentW last updated on 09/Apr/23 $${if}\:\:{x}^{\mathrm{2}} +\left({m}−\mathrm{2}\right){x}+\mathrm{2}{m}=\mathrm{0}\:\:{and} \\ $$$$\left({x}_{\mathrm{1}} −\mathrm{1}\right)\left({x}_{\mathrm{2}} −\mathrm{1}\right)=\mathrm{1}\:\:{then}\:{find}\:{the}\:{value} \\ $$$${of}\:\:\:{m}=? \\ $$ Answered by mahdipoor last updated on…
Question Number 190716 by sciencestudentW last updated on 09/Apr/23 $${if}\:{x}+\mathrm{2}+\frac{{m}+\mathrm{2}}{{m}−\mathrm{2}}=\mathrm{1}\:\:{and}\:\:\:{x}_{\mathrm{1}} \centerdot{x}_{\mathrm{2}} =\mathrm{2}\:\:{then} \\ $$$${find}\:{the}\:{value}\:{of}\:\:{m}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 190715 by sciencestudentW last updated on 09/Apr/23 $${if}\:\left(\mathrm{5}{m}−\mathrm{1}\right){x}^{\mathrm{2}} −\left(\mathrm{5}{m}+\mathrm{2}\right){x}+\mathrm{3}{m}−\mathrm{2}=\mathrm{0} \\ $$$${and}\:\:{x}_{\mathrm{1}} ={x}_{\mathrm{2}} \:\:{then}\:{find}\:\:{m}=? \\ $$ Answered by cortano12 last updated on 10/Apr/23 $$\:\Rightarrow\:\mathrm{25m}^{\mathrm{2}}…
Question Number 190719 by 07049753053 last updated on 09/Apr/23 $$\boldsymbol{\mathrm{F}{ind}}\:\boldsymbol{{the}}\:\boldsymbol{{real}}\:\boldsymbol{{part}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{\mathrm{R}{e}}\left(\frac{\boldsymbol{\Gamma}\left({a}+\mathrm{1}\right)}{\left(\mathrm{1}−{a}\right)^{{a}+\mathrm{1}} }\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 125173 by mohammad17 last updated on 08/Dec/20 Commented by mohammad17 last updated on 08/Dec/20 $${please}\:{help}\:{me} \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 125165 by Mammadli last updated on 08/Dec/20 $$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{sinx}}^{\boldsymbol{{cosx}}} \\ $$$$\boldsymbol{{f}}\:'\left(\boldsymbol{{x}}\right)=? \\ $$ Answered by Dwaipayan Shikari last updated on 08/Dec/20 $${f}\left({x}\right)=\left({sinx}\right)^{{cosx}} \\ $$$${log}\left({f}\left({x}\right)\right)={cosxlog}\left({sinx}\right)…
Question Number 59626 by naka3546 last updated on 12/May/19 $${Rationalize}\:\:{the}\:\:{denominator}\:\:{of} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{1}\:−\:\sqrt{\mathrm{2}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}}} \\ $$ Answered by MJS last updated on 12/May/19 $$\mathrm{first}\:\mathrm{use}\:\mathrm{this}\:\mathrm{formula} \\ $$$$\frac{\mathrm{1}}{{a}−\sqrt{{b}}}=\frac{{a}+\sqrt{{b}}}{\left({a}−\sqrt{{b}}\right)\left({a}+\sqrt{{b}}\right)}=\frac{{a}+\sqrt{{b}}}{{a}^{\mathrm{2}} −{b}}…
Question Number 59627 by naka3546 last updated on 12/May/19 Answered by tanmay last updated on 12/May/19 $$\frac{\mathrm{3}−\mathrm{1}}{\mathrm{3}}\left[\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\right)\right] \\ $$$$=\left(\frac{{b}+{c}}{{a}^{\mathrm{2}} }+\frac{{c}+{a}}{{b}^{\mathrm{2}} }+\frac{{a}+{b}}{{c}^{\mathrm{2}} }+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)−\frac{\mathrm{1}}{\mathrm{3}}\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}}…
Question Number 59624 by naka3546 last updated on 12/May/19 $${Rationalize}\:\:\:{the}\:\:{denominator}\:\:{of} \\ $$$$\:\:\:\:\:\:\frac{\mathrm{2}}{\:\sqrt{{x}+\mathrm{2}}\:\:+\:\:\sqrt{{x}+\mathrm{1}}\:\:+\:\:\sqrt{{x}}} \\ $$ Answered by MJS last updated on 12/May/19 $$\frac{\mathrm{1}}{\:\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}}=\frac{\sqrt{{a}}−\sqrt{{b}}−\sqrt{{c}}}{{a}−{b}−{c}−\mathrm{2}\sqrt{{bc}}}= \\ $$$$=\frac{\left(\sqrt{{a}}−\sqrt{{b}}−\sqrt{{c}}\right)\left({a}−{b}−{c}+\mathrm{2}\sqrt{{bc}}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}}…