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 $${a}^{\mathrm{2}} +{a}=\mathrm{5} \\ $$$${Let}\:{u}={a}+\mathrm{3}\Rightarrow{a}={u}−\mathrm{3} \\ $$$$\Rightarrow\left({u}−\mathrm{3}\right)^{\mathrm{2}} +{u}−\mathrm{3}=\mathrm{5} \\ $$$${u}^{\mathrm{2}}…
Question Number 7076 by Tawakalitu. last updated on 09/Aug/16 Commented by Yozzii last updated on 09/Aug/16 $$\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{1}/\mathrm{3}} =\left\{\sqrt{\mathrm{5}}\left(\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)\right\}^{\mathrm{1}/\mathrm{3}} =\mathrm{5}^{\mathrm{1}/\mathrm{6}} \left(\mathrm{1}+{u}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${u}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}<\mathrm{1}. \\ $$$${Also}\:\left(\mathrm{2}−\sqrt{\mathrm{5}}\right)^{\mathrm{1}/\mathrm{3}} =−\mathrm{5}^{\mathrm{1}/\mathrm{6}}…
Question Number 72608 by aliesam last updated on 30/Oct/19 Answered by mind is power last updated on 30/Oct/19 $$\mathrm{s}\left(\mathrm{x}\right)=\mathrm{n}\Rightarrow\mathrm{x}\in\left[\mathrm{10}^{\mathrm{n}−\mathrm{1}} ,\mathrm{10}^{\mathrm{n}} \left[\right.\right. \\ $$$$\mathrm{s}\left(\mathrm{y}\right)=\mathrm{m}\Rightarrow\mathrm{y}\in\left[\mathrm{10}^{\mathrm{m}−\mathrm{1}} ,\mathrm{10}^{\mathrm{m}} \left[\right.\right.…
Question Number 7075 by Tawakalitu. last updated on 09/Aug/16 $${Find}\:{all}\:{the}\:{solution}\:{for}\:{x}\:{and}\:{y}\:{in}\:{the}\:{equation} \\ $$$$ \\ $$$${x}^{{y}} \:+\:{y}^{{x}} \:=\:\mathrm{17}\:\:\:\:……..\:{equation}\:\left({i}\right) \\ $$$${x}\:+\:{y}\:=\:\mathrm{5}\:\:\:\:\:\:………\:{equation}\:\left({ii}\right) \\ $$ Terms of Service Privacy Policy…