Question Number 129979 by liberty last updated on 21/Jan/21 $$\mathrm{If}\:\begin{cases}{\mid\mathrm{x}\mid\:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{5}}\\{\mathrm{x}\:+\:\mid\mathrm{y}\mid\:−\mathrm{y}\:=\:\mathrm{10}}\end{cases}\:\mathrm{then}\:\mathrm{x}+\mathrm{y}\:? \\ $$ Answered by MJS_new last updated on 21/Jan/21 $$\mid{x}\mid+{x}+{y}=\mathrm{5} \\ $$$${x}\leqslant\mathrm{0}\:\Rightarrow\:{y}=\mathrm{5} \\ $$$$\mathrm{but}\:\mathrm{then}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{eq}.\:\mathrm{gives}\:{x}=\mathrm{10}>\mathrm{0}…
Question Number 64404 by aliesam last updated on 17/Jul/19 Commented by Prithwish sen last updated on 17/Jul/19 $$\mathrm{x}+\mathrm{1}\:=\:\sqrt{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \:}\:=\:\sqrt{\mathrm{1}+\mathrm{x}\left(\mathrm{x}+\mathrm{2}\right)}\:=\:\sqrt{\mathrm{1}+\mathrm{x}\sqrt{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} }} \\ $$$$=\:\sqrt{\mathrm{1}+\mathrm{x}\sqrt{\mathrm{1}+\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{3}\right)}\:}\:=\:\sqrt{\mathrm{1}+\mathrm{x}\sqrt{\mathrm{1}+\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} }}} \\ $$$$=\:\sqrt{\mathrm{1}+\mathrm{x}\sqrt{\mathrm{1}+\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{1}+\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{4}\right)}}\:}…
Question Number 64378 by aliesam last updated on 17/Jul/19 Commented by Prithwish sen last updated on 17/Jul/19 $$\mathrm{Wow}! \\ $$ Terms of Service Privacy Policy…
Question Number 129841 by Adel last updated on 20/Jan/21 $$\left\{_{\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{2x}} }+\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{4x}} }+\mathrm{4}^{\mathrm{x}} =\mathrm{5}} ^{\sqrt[{\mathrm{3}}]{\mathrm{2}^{\mathrm{x}} \:}−\mathrm{2}^{\mathrm{x}} =\mathrm{2}} \right\}\Rightarrow\mathrm{2}^{\mathrm{x}} −\mathrm{8}^{\mathrm{x}} =? \\ $$$$\mathrm{pleas}\:\mathrm{solve}\:\mathrm{thes} \\ $$ Commented by…
Question Number 129832 by ajfour last updated on 19/Jan/21 Commented by ajfour last updated on 19/Jan/21 $${If}\:{p},\:{q},\:{r}\:{are}\:{roots}\:{of}\:{y}={x}^{\mathrm{3}} −{x}−{c} \\ $$$${then}\:{find}\:{p}\:{or}\:{q}\:{or}\:{r},\:{with}\:{the} \\ $$$${help}\:{of}\:{s}. \\ $$ Commented…
Question Number 64284 by Tawa1 last updated on 16/Jul/19 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{integer}\:\mathrm{solution}\:\:\:\:\:\:\:\mathrm{3x}^{\mathrm{2}} \:+\:\mathrm{1}\:\:=\:\:\mathrm{4y}^{\mathrm{3}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 64187 by Hope last updated on 15/Jul/19 Answered by Hope last updated on 15/Jul/19 $$\frac{{yz}}{{xyz}}+\frac{\mathrm{4}{zx}}{{xyz}}+\frac{\mathrm{9}{xy}}{{xyz}} \\ $$$$\frac{\mathrm{1}^{\mathrm{2}} }{{x}}+\frac{\mathrm{2}^{\mathrm{2}} }{{y}}+\frac{\mathrm{3}^{\mathrm{2}} }{{z}}\geqslant\frac{\left(\mathrm{1}+\mathrm{2}+\mathrm{3}\right)^{\mathrm{2}} }{\left({x}+{y}+{z}\right)} \\ $$$$\frac{\mathrm{1}^{\mathrm{2}}…
Question Number 64186 by Hope last updated on 15/Jul/19 Answered by Hope last updated on 15/Jul/19 $$\frac{\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }{{x}+{y}}+\frac{\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }{{y}+{z}}+\frac{\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }{{z}+{x}}\geqslant\frac{\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} }{\mathrm{2}\left({x}+{y}+{z}\right)} \\ $$$${do}\geqslant\frac{\mathrm{9}×\mathrm{2}}{\mathrm{2}\left({x}+{y}+{z}\right)} \\ $$$$\frac{\mathrm{2}}{{x}+{y}}+\frac{\mathrm{2}}{{y}+{z}}+\frac{\mathrm{2}}{{z}+{x}}\geqslant\frac{\mathrm{9}}{{x}+{y}+{z}}\:{proved}…
Question Number 64185 by Hope last updated on 15/Jul/19 Commented by Hope last updated on 15/Jul/19 $${all}\:{question}\:{based}\:{on}\:{Tito}−{lemma}\:{inequality} \\ $$ Commented by Tony Lin last updated…
Question Number 64174 by gajrajgchouhan last updated on 15/Jul/19 $$\mathrm{2}^{{n}} \:/\:\mathrm{5}^{\mathrm{2}^{{n}} } \:+\:\mathrm{1}\:{infinite}\:{series}\:{sum}\:{ffom}\:\mathrm{0}\:{to}\: \\ $$$${infinity} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com