Question Number 7147 by Tawakalitu. last updated on 13/Aug/16 Commented by Yozzii last updated on 13/Aug/16 $${u}\left({n}\right)=\left(−\mathrm{1}+\mathrm{2}\right)+\left(−\mathrm{3}+\mathrm{4}\right)+\left(−\mathrm{5}+\mathrm{6}\right)+…+\left(−\left(\mathrm{2}{n}−\mathrm{1}\right)+\mathrm{2}{n}\right) \\ $$$${u}\left({n}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{2}{n}+\mathrm{1}+\mathrm{2}{n}\right)={n} \\ $$$${u}\left(\frac{\mathrm{10}^{\mathrm{9}} }{\mathrm{2}}\right)=\frac{\mathrm{10}^{\mathrm{9}} }{\mathrm{2}}…
Question Number 7138 by Tawakalitu. last updated on 12/Aug/16 Answered by Yozzii last updated on 13/Aug/16 $${x}^{{x}^{\mathrm{2}} } =\mathrm{3}\:\:\left({x}>\mathrm{0}\right) \\ $$$${Let}\:{x}^{\mathrm{2}} ={e}^{{k}} \Rightarrow{x}={e}^{{k}/\mathrm{2}} \\ $$$$\therefore{e}^{\left({k}/\mathrm{2}\right){e}^{{k}}…
Question Number 7139 by Tawakalitu. last updated on 12/Aug/16 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 7127 by Tawakalitu. last updated on 11/Aug/16 Commented by Yozzii last updated on 11/Aug/16 $${Let}\:{u}={x}\sqrt{\mathrm{1}+\left({x}+\mathrm{1}\right)+\sqrt{\mathrm{1}+\left({x}+\mathrm{2}\right)+\sqrt{\mathrm{1}+\left({x}+\mathrm{3}\right)+\sqrt{\mathrm{1}+….}}}} \\ $$$$\Rightarrow\frac{{u}}{{x}}=\sqrt{\mathrm{1}+\left({x}+\mathrm{1}\right)+\sqrt{\mathrm{1}+\left({x}+\mathrm{2}\right)+\sqrt{\mathrm{1}+\left({x}+\mathrm{3}\right)+\sqrt{\mathrm{1}+….}}}} \\ $$$${let}\:{n}={x}+\mathrm{1}\Rightarrow{x}={n}−\mathrm{1}. \\ $$$$\therefore\frac{{u}}{{n}−\mathrm{1}}=\sqrt{\mathrm{1}+{n}+\sqrt{\mathrm{1}+\left({n}+\mathrm{1}\right)+\sqrt{\mathrm{1}+\left({n}+\mathrm{2}\right)+\sqrt{\mathrm{1}+\left({n}+\mathrm{3}\right)+\sqrt{\mathrm{1}+…..}}}}} \\ $$$$\frac{{u}^{\mathrm{2}}…
Question Number 7125 by Tawakalitu. last updated on 11/Aug/16 Commented by Tawakalitu. last updated on 11/Aug/16 $${I}\:{have}\:{seen}\:{a}\:{solution}\:{to}\:{this}\:{question} \\ $$ Answered by sandy_suhendra last updated on…
Question Number 7126 by Tawakalitu. last updated on 11/Aug/16 Commented by Tawakalitu. last updated on 11/Aug/16 $${I}\:{need}\:{solution}\:{to}\:{this}\:{one}.\:{thanks}\:{in}\:{advance} \\ $$ Commented by Yozzii last updated on…
Question Number 7087 by Tawakalitu. last updated on 10/Aug/16 Commented by Yozzii last updated on 10/Aug/16 $${W}=\mathrm{2}^{\mathrm{129}} \mathrm{3}^{\mathrm{81}} \mathrm{5}^{\mathrm{131}} \\ $$$${X}=\mathrm{2}^{\mathrm{127}} \mathrm{3}^{\mathrm{81}} \mathrm{5}^{\mathrm{131}} \\ $$$${Y}=\mathrm{2}^{\mathrm{126}}…
Question Number 138154 by JulioCesar last updated on 10/Apr/21 Answered by TheSupreme last updated on 10/Apr/21 $$\frac{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{x}}{{x}}\geqslant\left(\mathrm{1}−{x}\right)^{\mathrm{3}} \\ $$$$\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{{x}}\geqslant\left(\mathrm{1}−{x}\right)^{\mathrm{3}} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left[\frac{\mathrm{1}}{{x}}−\left(\mathrm{1}−{x}\right)\right]\geqslant\mathrm{0} \\…
Question Number 138150 by liberty last updated on 10/Apr/21 $${Given}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left({y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\right)=\mathrm{9} \\ $$$${find}\:{the}\:{value}\:{of}\: \\ $$$$\:\:\:\:\:\:\:{x}\sqrt{{y}^{\mathrm{2}} +\mathrm{4}}\:+\:{y}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:. \\ $$ Answered by EDWIN88 last updated…
Question Number 7077 by Tawakalitu. last updated on 09/Aug/16 Commented by Yozzii last updated on 09/Aug/16 $${a}^{\mathrm{2}} +{a}=\mathrm{5} \\ $$$${Let}\:{u}={a}+\mathrm{3}\Rightarrow{a}={u}−\mathrm{3} \\ $$$$\Rightarrow\left({u}−\mathrm{3}\right)^{\mathrm{2}} +{u}−\mathrm{3}=\mathrm{5} \\ $$$${u}^{\mathrm{2}}…