Question Number 171742 by Mikenice last updated on 20/Jun/22 $${solve}:\:\left(\mathrm{1}+{n}\right)!+\mathrm{2}\left({n}\right)!=\left({n}+\mathrm{2}\right)!−\mathrm{4}{n}! \\ $$ Commented by kaivan.ahmadi last updated on 20/Jun/22 $$\left({n}+\mathrm{2}\right)!−\left({n}+\mathrm{1}\right)!−\mathrm{6}{n}!=\mathrm{0} \\ $$$${n}!\left[\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)−\left({n}+\mathrm{1}\right)−\mathrm{6}\right]=\mathrm{0} \\ $$$$\Rightarrow{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}−{n}−\mathrm{1}−\mathrm{6}=\mathrm{0}\Rightarrow…
Question Number 171743 by Mikenice last updated on 20/Jun/22 $${solve}:\:\mathrm{12}^{{x}−\mathrm{2}} =\mathrm{4}^{{x}} ,\:{find}\:{x} \\ $$ Commented by kaivan.ahmadi last updated on 20/Jun/22 $$\mathrm{3}^{{x}−\mathrm{2}} ×\mathrm{4}^{{x}−\mathrm{2}} =\mathrm{4}^{{x}} \Rightarrow\mathrm{3}^{{x}−\mathrm{2}}…
Question Number 171739 by Mikenice last updated on 20/Jun/22 $${the}\:{result}\:{of}\:{dividing}\:\left(\frac{{x}^{{a}} }{{x}^{{b}} }\right)^{{a}+{b}} {by}\:\left(\frac{{x}^{{a}+{b}} }{{x}^{{a}−{b}} }\right)^{\frac{{a}^{\mathrm{2}} }{{b}\:}} {is} \\ $$ Commented by kaivan.ahmadi last updated on…
Question Number 171728 by Mikenice last updated on 20/Jun/22 $$\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }{{x}+{y}}=\mathrm{7} \\ $$$$\frac{{x}^{\mathrm{3}} −{y}^{\mathrm{3}} }{{x}−{y}}=\mathrm{19}.\:{find}\:{x}\:{and}\:{y} \\ $$ Answered by Ar Brandon last updated on…
Question Number 171730 by Mikenice last updated on 20/Jun/22 $${solve}\frac{\mathrm{8}^{{x}} −\mathrm{2}^{{x}} }{\mathrm{6}^{{x}} −\mathrm{3}^{{x}\:} }=\mathrm{2}.\:{find}\:{x} \\ $$$$ \\ $$ Answered by puissant last updated on 20/Jun/22…
Question Number 106188 by mr W last updated on 03/Aug/20 $${Solve}\:{for}\:{x} \\ $$$${x}^{{x}^{…{x}^{{a}} } } ={a}\:{with}\:{a}\in\mathbb{R}^{+} \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 106176 by john santu last updated on 03/Aug/20 $$\mathrm{If}\:\mathrm{20}\:\mathrm{men}\:\mathrm{can}\:\mathrm{lay}\:\mathrm{36m}\:\mathrm{of}\:\mathrm{a}\:\mathrm{pipe} \\ $$$$\mathrm{in}\:\mathrm{8}\:\mathrm{hours}.\:\mathrm{How}\:\mathrm{long}\:\mathrm{would}\:\mathrm{25} \\ $$$$\mathrm{men}\:\mathrm{take}\:\mathrm{to}\:\mathrm{lay}\:\mathrm{the}\:\mathrm{next}\:\mathrm{54m}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{pipe}? \\ $$ Commented by john santu last updated…
Question Number 106156 by bemath last updated on 03/Aug/20 $$\mathrm{Given}\:\begin{cases}{\sqrt{\mathrm{xy}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{y}}}\:=\mathrm{9}}\\{\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\:=\:\mathrm{20}}\end{cases} \\ $$$$\mathrm{where}\:\mathrm{x}\:>\:\mathrm{y}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{x}\sqrt{\mathrm{y}}\:−\mathrm{y}\sqrt{\mathrm{x}}\:. \\ $$ Commented by bemath last updated on 03/Aug/20 $$\mathrm{let}\:\sqrt{\mathrm{x}}\:=\:\mathrm{a}\:\&\:\sqrt{\mathrm{y}}\:=\:\flat\: \\…
Question Number 171688 by Shrinava last updated on 20/Jun/22 $$\mathrm{In}\:\:\bigtriangleup\mathrm{ABC} \\ $$$$\mathrm{AA}^{'} \:,\:\mathrm{BB}^{'} \:,\:\mathrm{CC}^{'} \:-\:\mathrm{cevians} \\ $$$$\mathrm{AA}^{'} \:\cap\:\mathrm{BB}^{'} \:\cap\:\mathrm{CC}^{'} \:=\:\left\{\mathrm{P}\right\} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{min}\left(\left[\mathrm{APC}^{'} \right],\left[\mathrm{BPA}^{'}…
Question Number 171681 by Shrinava last updated on 19/Jun/22 Answered by mindispower last updated on 19/Jun/22 $${We}\:{Have}\:{Ln}\:{increase}\:{function}\:\:{witch}\:{is}\:{octave}\:{function} \\ $$$$\Leftrightarrow{aln}\left({cot}\left({A}\right)\right)+{bln}\left({cot}\left({B}\right)\right)+{cln}\left({cot}\left({C}\right)\right)\leqslant\left({a}+{b}+{c}\right){ln}\left(\frac{\mathrm{3}{R}}{{a}+{b}+{c}}\right) \\ $$$$\frac{\mathrm{1}}{{a}+{b}+{c}}\left({aln}\left({cot}\left({A}\right)\right)+{bln}\left({Cot}\left({B}\right)\right)+{cln}\left({cot}\left({C}\right)\right)\leqslant\right. \\ $$$${ln}\left(\frac{{acot}\left({A}\right)}{{a}+{b}+{c}}+\frac{{bcot}\left({B}\right)}{{a}+{b}+{c}}+\frac{{cCot}\left({C}\right)}{{a}+{b}+{c}}\right)….{E} \\ $$$$={ln}\left(\frac{\mathrm{1}}{{a}+{b}+{c}}\left(\frac{{acos}\left({A}\right)}{{sin}\left({A}\right)}+\frac{{bCos}\left({B}\right)}{{sin}\left({B}\right)}+\frac{{cCos}\left({C}\right)}{{sin}\left({C}\right)}\right)\right.…