Question Number 13223 by Tinkutara last updated on 16/May/17 $$\mathrm{If}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\mathrm{1}\:+\:\frac{{b}−{c}}{{a}}\right)^{{a}} \left(\mathrm{1}\:+\:\frac{{c}−{a}}{{b}}\right)^{{b}} \left(\mathrm{1}\:+\:\frac{{a}−{b}}{{c}}\right)^{{c}} \:<\:\mathrm{1} \\ $$ Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/May/17 $${if}:\:\:{a}={b}={c}\Rightarrow{RHS}=\mathrm{1}\nless{LHS}=\mathrm{1}…
Question Number 78755 by TawaTawa last updated on 20/Jan/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\:\:\:\:\mathrm{xy}\:+\:\mathrm{5x}\:+\:\mathrm{5y}\:\:=\:\:−\:\mathrm{25}\:\:\:\:\:\:…\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\mathrm{yz}\:+\:\mathrm{3y}\:+\:\mathrm{5z}\:\:=\:\:−\:\mathrm{15}\:\:\:\:\:\:…\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\mathrm{xz}\:+\:\mathrm{5z}\:+\:\mathrm{3x}\:\:=\:\:−\:\mathrm{15}\:\:\:\:\:\:…\:\left(\mathrm{iii}\right) \\ $$ Commented by Tony Lin last updated on…
Question Number 144264 by loveineq last updated on 24/Jun/21 $$\mathrm{Let}\:{a},{b}\:>\:\mathrm{0}\:\mathrm{and}\:\mathrm{2}{a}+{b}\:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{the}\:\mathrm{followings}:\:\:\:\:\:\:\:\:\:\: \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{{n}}{a}\left({b}+\mathrm{4}\right)+\mathrm{3}{b}^{\frac{\mathrm{1}}{{n}}} \:\leqslant\:\frac{\mathrm{10}+\mathrm{3}{n}}{{n}},\:\forall{n}\in\mathbb{N}^{+} \geqslant\mathrm{1}. \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{na}\left({b}+\mathrm{4}\right)+\mathrm{3}{b}^{{n}} \:\geqslant\:\mathrm{10}{n}+\mathrm{3},\:\forall{n}\in\mathbb{N}^{+} \geqslant\mathrm{2}. \\ $$$$ \\ $$ Terms of Service…
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…