Question Number 144270 by enter last updated on 24/Jun/21 Answered by ArielVyny last updated on 24/Jun/21
Question Number 78732 by Pratah last updated on 20/Jan/20 Commented by john santu last updated on 20/Jan/20 $$\left(\mathrm{1}\right)\:\left(\mathrm{6}−{y}\right)^{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{6}}} \:=\:\mathrm{3}^{{x}} \\ $$$$\left(\mathrm{2}\right)\:\left(\mathrm{6}−{z}\right)^{\sqrt{{y}^{\mathrm{2}} +\mathrm{2}{y}+\mathrm{6}}} \:=\:\mathrm{3}^{{y}} \\…
Question Number 144249 by mathdanisur last updated on 23/Jun/21 $$\left(\sqrt{\mathrm{5}\:+\:\mathrm{2}\sqrt{\mathrm{6}}}\right)^{\boldsymbol{{z}}} \:+\:\left(\sqrt{\mathrm{5}\:-\:\mathrm{2}\sqrt{\mathrm{6}}}\right)^{\boldsymbol{{z}}} \:=\:\mathrm{10} \\ $$$${Find}:\:\boldsymbol{{z}}=? \\ $$ Answered by Dwaipayan Shikari last updated on 23/Jun/21 $$\mathrm{2}…
Question Number 144237 by mathdanisur last updated on 23/Jun/21 Answered by MJS_new last updated on 23/Jun/21 $${t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}} \\ $$$$\mathrm{3}×\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}−\mathrm{4}×\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{3} \\ $$$${t}^{\mathrm{2}} −\frac{\mathrm{6}}{\mathrm{7}}{t}−\frac{\mathrm{1}}{\mathrm{7}}=\mathrm{0}…
Question Number 13166 by tawa tawa last updated on 15/May/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\:\mathrm{x}\:\:\mathrm{and}\:\:\mathrm{y} \\ $$$$ \\ $$$$\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{2xy}\:−\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{14}\:\:\:\:\:\:\:\:\:…………\:\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{2x}^{\mathrm{2}} \:+\:\mathrm{3xy}\:+\:\mathrm{y}^{\mathrm{2}} \:=\:−\:\mathrm{2}\:\:\:\:\:…………\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$ Answered by…
Question Number 144234 by mathdanisur last updated on 23/Jun/21 $$\int\:\frac{\sqrt{{x}^{\mathrm{2}} \:-\:{x}}}{{x}^{\mathrm{3}} }\:{dx}\:=\:? \\ $$ Answered by liberty last updated on 23/Jun/21 $$\int\:\frac{\sqrt{\mathrm{1}−\mathrm{x}^{−\mathrm{1}} }}{\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:\int\:\mathrm{x}^{−\mathrm{2}} \:\sqrt{\mathrm{1}−\mathrm{x}^{−\mathrm{1}}…
Question Number 78694 by TawaTawa last updated on 19/Jan/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}. \\ $$$$\:\:\:\:\:\mathrm{x}^{\mathrm{2}} \:−\:\left(\mathrm{y}\:−\:\mathrm{z}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{10}\:\:\:\:\:\:…\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\mathrm{y}^{\mathrm{2}} \:−\:\left(\mathrm{z}\:−\:\mathrm{x}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{5}\:\:\:\:\:\:…\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\mathrm{z}^{\mathrm{2}} \:−\:\left(\mathrm{x}\:\:−\:\mathrm{y}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{2}\:\:\:\:\:\:…\:\left(\mathrm{iii}\right) \\ $$ Commented…
Question Number 13155 by Tinkutara last updated on 15/May/17 Answered by mrW1 last updated on 16/May/17 $$=\left(\frac{{x}+{y}}{{x}}\right)\left(\frac{{y}+{z}}{{y}}\right)\left(\frac{{z}+{x}}{{z}}\right) \\ $$$$=\left(\mathrm{1}+\frac{{y}}{{x}}\right)\left(\mathrm{1}+\frac{{z}}{{y}}\right)\left(\mathrm{1}+\frac{{x}}{{z}}\right) \\ $$$$\geqslant\mathrm{2}\sqrt{\frac{{y}}{{x}}}×\mathrm{2}\sqrt{\frac{{z}}{{y}}}×\mathrm{2}\sqrt{\frac{{x}}{{z}}}=\mathrm{8} \\ $$$$ \\ $$$${x}+{y}+{z}=\mathrm{1}\:{is}\:{not}\:{necessary}!…
Question Number 13153 by Tinkutara last updated on 15/May/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/May/17 $$\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)\left(\mathrm{1}+{d}\right)= \\ $$$$\left({abcd}+{a}\right)\left({abcd}+{b}\right)\left({abcd}+{c}\right)\left({abcd}+{d}\right)= \\ $$$${abcd}\left({dbc}+\mathrm{1}\right)\left({acd}+\mathrm{1}\right)\left({abd}+\mathrm{1}\right)\left({abc}+\mathrm{1}\right)= \\ $$$$\left({abc}+\mathrm{1}\right)\left({abd}+\mathrm{1}\right)\left({acd}+\mathrm{1}\right)\left({dbc}+\mathrm{1}\right)\geqslant \\ $$$$\geqslant\mathrm{2}\sqrt{{abc}×\mathrm{1}}×\mathrm{2}\sqrt{{abd}×\mathrm{1}}×\mathrm{2}\sqrt{{acd}×\mathrm{1}}×\mathrm{2}\sqrt{{dbc}×\mathrm{1}}=…
Question Number 13145 by Joel577 last updated on 15/May/17 $$\sqrt[{\sqrt[{\sqrt{\mathrm{5}}}]{\mathrm{3}}}]{\mathrm{64}}\:=\:{x} \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{write}\:{x}\:\mathrm{into}\:\mathrm{fraction}\:\mathrm{exponent}\:\mathrm{form}? \\ $$ Answered by mrW1 last updated on 16/May/17 $${x}=\mathrm{64}^{\frac{\mathrm{1}}{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{2}}} }} }} \\…