Question Number 148268 by mathdanisur last updated on 26/Jul/21 Answered by Olaf_Thorendsen last updated on 26/Jul/21 $${z}^{\mathrm{4}} −\mathrm{3}{z}^{\mathrm{2}} +\mathrm{1}\:=\:\sqrt{\frac{\mathrm{4}}{\mathrm{4}−{z}^{\mathrm{2}} }−\mathrm{1}}\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{w}\:=\:{z}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\::\:{w}^{\mathrm{2}} −\mathrm{3}{w}+\mathrm{1}\:=\:\sqrt{\frac{\mathrm{4}}{\mathrm{4}−{w}}−\mathrm{1}}…
Question Number 148262 by mathdanisur last updated on 26/Jul/21 $${if}\:\:{x}>−\mathrm{1}\:\:;\:\:\left[{q}\right]={n}\geqslant\mathrm{2}\:\:;\:\:\left[\ast\right]-{GIF}\:\:{then}: \\ $$$$\left(\mathrm{1}+{x}\right)^{\boldsymbol{{q}}} \:\geqslant\:\left(\mathrm{1}+{nx}\right)\left(\mathrm{1}+\left({q}-{n}\right){x}\right)\:\geqslant\:\mathrm{1}+{qx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 148249 by mnjuly1970 last updated on 26/Jul/21 Commented by Kamel last updated on 26/Jul/21 $$ \\ $$$$\Omega_{{n}} =\int_{\mathrm{0}} ^{+\infty} \frac{{cos}\left(\pi{x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}}…
Question Number 82701 by jagoll last updated on 23/Feb/20 $$\mathrm{2}{f}\left({x}−\mathrm{1}\right)\:+\mathrm{3}{f}\left({x}+\mathrm{1}\right)\overset{\:} {\:}=\:\mathrm{3}{x}^{\mathrm{2}} −\mathrm{5}{x} \\ $$$${find}\:{f}\left({x}\right) \\ $$ Answered by TANMAY PANACEA last updated on 23/Feb/20 $${f}\left({x}\right)={ax}^{\mathrm{2}}…
Question Number 148229 by iloveisrael last updated on 26/Jul/21 Commented by nimnim last updated on 26/Jul/21 $$\:\:\:\:\:\:\:\:\:{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}=\mathrm{20} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} −\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}}{{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}×\frac{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}}…
Question Number 148226 by mathdanisur last updated on 26/Jul/21 Answered by mitica last updated on 26/Jul/21 $$\Sigma\frac{\mathrm{1}}{{x}\left({px}+\mathrm{1}\right)}=\Sigma\frac{\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} }{{p}+\frac{\mathrm{1}}{{x}}}\geqslant \\ $$$$\frac{\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} }{{p}+{q}+{r}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}}=\frac{\mathrm{9}}{{p}+{q}+{r}+\mathrm{3}} \\ $$ Commented by…
Question Number 148203 by liberty last updated on 26/Jul/21 $$\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{8}}{{n}^{\mathrm{2}} +{n}}\right)=? \\ $$ Answered by mathmax by abdo last updated on 26/Jul/21 $$\mathrm{S}=\mathrm{8}\sum_{\mathrm{n}=\mathrm{1}}…
Question Number 82667 by M±th+et£s last updated on 23/Feb/20 $${if}\:{a}>\mathrm{0}\:\:{b}>\mathrm{0}\:\:{a}\leqslant{b} \\ $$$$ \\ $$$${show}\:{that}\: \\ $$$${a}^{\mathrm{2}} \leqslant\left(\frac{\mathrm{2}{ab}}{{a}+{b}}\right)^{\mathrm{2}} \leqslant{ab}\leqslant\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}\leqslant{b}^{\mathrm{2}} \\ $$ Answered by…
Question Number 82665 by jagoll last updated on 23/Feb/20 $$\sqrt{\mathrm{6561}\sqrt{\mathrm{6561}\sqrt{\mathrm{6561}\sqrt{\mathrm{6561}\sqrt{…}}}}}\:=\:\mathrm{3}^{\:\mathrm{8}^{{x}} \:} \\ $$$$\sqrt{\mathrm{6561}\sqrt{…}}\:\:\left(\mathrm{60}\:{time}\right) \\ $$$${find}\:{x} \\ $$ Commented by mr W last updated on 23/Feb/20…
Question Number 82644 by Power last updated on 23/Feb/20 Commented by Power last updated on 23/Feb/20 $$\mathrm{sorry}\:\:\mathrm{it}\:\mathrm{worked} \\ $$ Commented by john santu last updated…