Question Number 146914 by mathdanisur last updated on 16/Jul/21 $$\mid{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\mid\:=\:\mid{x}−\mathrm{4}\mid\:\Rightarrow\:{x}=? \\ $$ Answered by liberty last updated on 16/Jul/21 $$\:\mid{a}\mid=\mid{b}\mid\:\Leftrightarrow\:\left({a}+{b}\right)\left({a}−{b}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}+{x}−\mathrm{4}\right)\left({x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}−{x}+\mathrm{4}\right)=\mathrm{0}…
Question Number 146913 by mathdanisur last updated on 16/Jul/21 $${ax}={by}={cz}=\frac{\mathrm{2}}{\mathrm{3}}\:{and}\:{ab}+{bc}+{ac}=\mathrm{36}{abc} \\ $$$${find}\:\:{x}+{y}+{z}=? \\ $$ Answered by liberty last updated on 16/Jul/21 $$\:\begin{cases}{{ax}=\frac{\mathrm{2}}{\mathrm{3}}}\\{{by}=\frac{\mathrm{2}}{\mathrm{3}}}\\{{cz}=\frac{\mathrm{2}}{\mathrm{3}}}\end{cases}\: \\ $$$$\Rightarrow{x}+{y}+{z}=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right) \\…
Question Number 81369 by mr W last updated on 12/Feb/20 Commented by mr W last updated on 12/Feb/20 $${This}\:{is}\:{a}\:{repost}\:{of}\:{Q}\mathrm{81308}. \\ $$$$ \\ $$$${I}\:{found}\:{the}\:{answer}\:{through}\:{the} \\ $$$${following}\:{way},\:{but}\:{it}\:{seems}\:{lengthy}…
Question Number 146896 by mathdanisur last updated on 16/Jul/21 $$\frac{\mathrm{5}}{\:\sqrt[{\mathrm{8}}]{\mathrm{6}}\:+\:\mathrm{1}}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{6}}\:+\:\mathrm{1}}\:\centerdot\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}\:+\:\mathrm{1}}\:+\:\mathrm{1}\:=\:? \\ $$ Answered by EDWIN88 last updated on 16/Jul/21 $$\:\mathrm{What}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\:\frac{\mathrm{5}}{\:\sqrt[{\mathrm{8}}]{\mathrm{6}}\:+\mathrm{1}}\:×\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{6}}\:+\mathrm{1}}\:×\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}\:+\mathrm{1}}\:+\:\mathrm{1}\:. \\ $$$$\:\mathrm{Solution}\::\: \\…
Question Number 146895 by mathdanisur last updated on 16/Jul/21 $$\frac{\mathrm{1}}{\mathrm{2}\:+\:\boldsymbol{{log}}_{\mathrm{3}} \left(\mathrm{25}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{2}\:+\:\boldsymbol{{log}}_{\mathrm{5}} \left(\mathrm{9}\right)}\:=\:? \\ $$ Answered by EDWIN88 last updated on 16/Jul/21 $$\Rightarrow\:\mathrm{L}\:=\frac{\mathrm{1}}{\mathrm{2}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{25}\right)}+\frac{\mathrm{1}}{\mathrm{2}+\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{9}\right)} \\…
Question Number 146865 by mathdanisur last updated on 16/Jul/21 $${arcsin}\left({x}^{\mathrm{2}} −\mathrm{4}\right)\:=\:{arcsin}\left(\mathrm{2}{x}\:+\:\mathrm{4}\right) \\ $$$$\Rightarrow\:{x}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 16/Jul/21 $$\mathrm{arcsin}\left({x}^{\mathrm{2}} −\mathrm{4}\right)\:=\:\mathrm{arcsin}\left(\mathrm{2}{x}+\mathrm{4}\right) \\…
Question Number 146860 by mathdanisur last updated on 16/Jul/21 $$\int{sin}\left({x}\right)\:{cos}\left({x}\right)\:{dx}\:=\:? \\ $$ Answered by puissant last updated on 16/Jul/21 $$=\int\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{2x}\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\mathrm{2x}\right)\right]+\mathrm{k} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\left(\mathrm{2x}\right)+\mathrm{k}. \\…
Question Number 146866 by mathdanisur last updated on 16/Jul/21 $$\frac{{sin}^{\mathrm{6}} \boldsymbol{\alpha}\:+\:{cos}^{\mathrm{6}} \boldsymbol{\alpha}\:-\:\mathrm{1}}{{sin}^{\mathrm{4}} \boldsymbol{\alpha}\:-\:{sin}^{\mathrm{2}} \boldsymbol{\alpha}}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 16/Jul/21 $${a}\:=\:\mathrm{sin}^{\mathrm{6}} \alpha+\mathrm{cos}^{\mathrm{6}}…
Question Number 146859 by mathdanisur last updated on 16/Jul/21 $$\underset{\:\mathrm{0}} {\overset{\mathrm{3}} {\int}}\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{8}{x}}\:=\:? \\ $$ Answered by mathmax by abdo last updated on 16/Jul/21 $$\mathrm{I}=\int_{\mathrm{0}}…
Question Number 146854 by mathdanisur last updated on 16/Jul/21 $${Solve}\:{for}\:{real}\:{numbers}\:{the}\:{following} \\ $$$${system}\:{of}\:{equations} \\ $$$$\begin{cases}{{a}\left({a}+\mathrm{1}\right)\:=\:{b}−\mathrm{1}}\\{{a}^{\mathrm{2}} \left({b}+\mathrm{3}\right)+\mathrm{2}{a}\:=\:−\mathrm{1}}\end{cases} \\ $$ Commented by Mrsof last updated on 16/Jul/21 $${a}\left({a}+{b}\right)+\mathrm{4}={b}+\mathrm{3}…