Question Number 81301 by mr W last updated on 11/Feb/20 Commented by mr W last updated on 11/Feb/20 $${Find}\:{f}\left({x}\right)=? \\ $$$$\left({x}\in\mathbb{N}\right) \\ $$ Commented by…
Question Number 146817 by mathdanisur last updated on 15/Jul/21 Answered by mindispower last updated on 15/Jul/21 $$\Leftrightarrow\frac{{y}\left({x}+{y}\right)+{z}\left({x}+{z}\right)}{\:\sqrt{\left({x}+{y}\right)\left({x}+{z}\right)}}\geqslant{y}+{z}…. \\ $$$${y}\left({x}+{y}\right)+{z}\left({x}+{z}\right)={x}\left({y}+{z}\right)+{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \geqslant{x}\left({y}+{z}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left({y}+{z}\right)^{\mathrm{2}} \\ $$$$=\left({y}+{z}\right)\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\left({y}+{z}\right)\right)=\left({y}+{z}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\left({x}+{y}\right)+\left({x}+{z}\right)\right)\right) \\ $$$$\geqslant\left({y}+{z}\right).\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}\sqrt{\left({x}+{y}\right)\left({x}+{z}\right)}=\left({y}+{z}\right)\sqrt{\left({x}+{y}\right)\left({x}+{z}\right)}…
Question Number 15721 by ajfour last updated on 13/Jun/17 $${If}\:\:\:\:\boldsymbol{{x}}+\boldsymbol{{y}}+\boldsymbol{{z}}=\mathrm{1} \\ $$$$\:\:\:\:\:{with}\:\mathrm{0}<{x},\:{y},\:{z}\:<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:{then}\:{find}\:{tbe}\:\:{range}\:{of}\:{values}\:{of} \\ $$$$\:\:\:\frac{\mathrm{1}}{\boldsymbol{{x}}+\boldsymbol{{y}}}+\frac{\mathrm{1}}{\boldsymbol{{y}}+\boldsymbol{{z}}}+\frac{\mathrm{1}}{\boldsymbol{{z}}+\boldsymbol{{x}}}\:. \\ $$ Answered by mrW1 last updated on 13/Jun/17…
Question Number 81251 by mr W last updated on 10/Feb/20 Commented by mind is power last updated on 10/Feb/20 $${xyln}\left({x}\right)={ln}\left({y}\right) \\ $$$${xyln}\left({y}\right)=\mathrm{4}{ln}\left({x}\right) \\ $$$$\Leftrightarrow \\…
Question Number 146782 by mathdanisur last updated on 15/Jul/21 $${Find}\:{the}\:{modulus}\:{of}\:{a}\:{complex} \\ $$$${number}: \\ $$$${Z}\:=\:{cos}\:\mathrm{40}\:+\:{i}\:{sin}\:\mathrm{40}\:+\mathrm{1}\:=\:? \\ $$ Answered by Ar Brandon last updated on 15/Jul/21 $$\mathrm{z}=\mathrm{e}^{\mathrm{40i}}…
Question Number 146776 by mathdanisur last updated on 15/Jul/21 Commented by mr W last updated on 15/Jul/21 $${AB}^{\mathrm{2}} =\mathrm{20}^{\mathrm{2}} −{x}^{\mathrm{2}} =\left(\mathrm{20}+\mathrm{5}\right)^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} −\mathrm{2}×\mathrm{25}×\mathrm{10}\:\mathrm{cos}\:{C} \\ $$$${AD}^{\mathrm{2}}…
Question Number 81242 by panky0214 last updated on 10/Feb/20 Commented by mr W last updated on 10/Feb/20 $${i}\:{got}\:{maximum}=\frac{\mathrm{120}}{\mathrm{49}}\approx\mathrm{2}.\mathrm{45} \\ $$ Commented by mr W last…
Question Number 146778 by mathdanisur last updated on 15/Jul/21 $${Find}\:{the}\:{modulus}\:{of}\:{a}\:{complex} \\ $$$${number}: \\ $$$${Z}\:=\:{cos}\:\mathrm{40}\:+\:{i}\:{sin}\:\mathrm{20}\:+\:\mathrm{1}\:=\:? \\ $$ Answered by mathmax by abdo last updated on 15/Jul/21…
Question Number 146764 by Ar Brandon last updated on 15/Jul/21 $$\mathrm{A}=\left(\frac{\left(\mathrm{x}+\sqrt[{\mathrm{3}}]{\mathrm{2ax}^{\mathrm{2}} }\right)\left(\mathrm{2a}+\sqrt[{\mathrm{3}}]{\mathrm{2a}^{\mathrm{2}} \mathrm{x}}\right)^{−\mathrm{1}} −\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}}−\sqrt[{\mathrm{3}}]{\mathrm{2a}}}−\left(\mathrm{2a}\right)^{−\mathrm{1}/\mathrm{3}} \right)^{−\mathrm{6}} ,\:\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{R}^{\mathrm{2}} \\ $$$$\mathrm{a}-\:\mathrm{A}=\frac{\mathrm{16a}^{\mathrm{4}} }{\mathrm{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}-\:\mathrm{A}=\frac{\mathrm{8}}{\mathrm{ax}}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}-\mathrm{A}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{2a}}}{\mathrm{3x}^{\mathrm{3}} } \\ $$ Terms of…
Question Number 81223 by ubaydulla last updated on 10/Feb/20 $${x}\left(\mathrm{3}^{{x}} +\mathrm{2}\right)=\mathrm{3}\left(\mathrm{1}−\mathrm{3}^{{x}} \right)−{x}^{\mathrm{2}} \\ $$ Answered by MJS last updated on 10/Feb/20 $$\mathrm{3}^{{x}} {x}+\mathrm{2}{x}−\mathrm{3}+\mathrm{3}^{{x}} \mathrm{3}+{x}^{\mathrm{2}} =\mathrm{0}…