Question Number 133936 by liberty last updated on 25/Feb/21 $$\mathrm{log}\:_{\mathrm{x}+\mathrm{8}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{3x}−\mathrm{4}\right)\:<\:\mathrm{2}.\mathrm{log}\:_{\left(\mathrm{4}−\mathrm{x}\right)^{\mathrm{2}} } \left(\mid\mathrm{x}−\mathrm{4}\mid\right)\: \\ $$ Answered by EDWIN88 last updated on 25/Feb/21 $$\:\mathrm{log}\:_{\mathrm{x}+\mathrm{8}} \left(\mathrm{x}^{\mathrm{2}}…
Question Number 133938 by Algoritm last updated on 25/Feb/21 Answered by mathmax by abdo last updated on 25/Feb/21 $$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(−\mathrm{lnx}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\mathrm{changement}\:−\mathrm{lnx}=\mathrm{t}\:\mathrm{give}\:\mathrm{x}=\mathrm{e}^{−\mathrm{t}} \:\Rightarrow \\ $$$$\mathrm{I}=\int_{\mathrm{0}}…
Question Number 133934 by liberty last updated on 25/Feb/21 $$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{3}\right)^{\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{4}} \:\leqslant\:\mathrm{1} \\ $$ Answered by EDWIN88 last updated on 25/Feb/21 $$\left(\mathrm{i}\right)\:\mathrm{x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{3}\:>\:\mathrm{0}\:;\:\begin{cases}{\mathrm{x}<\mathrm{1}}\\{\mathrm{x}>\mathrm{3}}\end{cases} \\…
Question Number 68397 by Faradtimmy last updated on 10/Sep/19 Answered by $@ty@m123 last updated on 10/Sep/19 $$\left({a}\right)\:{LHS}=\sqrt{\mathrm{3}}\mathrm{cos}\:\theta−\mathrm{sin}\:\theta \\ $$$$=\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\theta\right) \\ $$$$=\mathrm{2}\left(\mathrm{cos}\:\mathrm{30cos}\:\theta−\mathrm{sin}\:\mathrm{30sin}\:\theta\right) \\ $$$$=\mathrm{2cos}\:\left(\mathrm{30}+\theta\right) \\ $$$${Pl}.\:{check}\:{the}\:{question}.…
Question Number 133929 by bemath last updated on 25/Feb/21 $$\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{76}°+\mathrm{cos}\:^{\mathrm{2}} \mathrm{16}°−\mathrm{cos}\:\mathrm{76}°.\mathrm{cos}\:\mathrm{16}°=? \\ $$ Answered by liberty last updated on 25/Feb/21 $$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}…
Question Number 2859 by Syaka last updated on 29/Nov/15 $${if}\:{x}^{\mathrm{2}} \:−\:\mathrm{2}{cx}\:−\:\mathrm{5}{d}\:=\:\mathrm{0}\:{has}\:{root}\:{a}\:{and}\:{b}, \\ $$$${also}\:{x}^{\mathrm{2}} \:−\:\mathrm{2}{ax}\:−\:\mathrm{5}{b}\:=\:\mathrm{0}\:{has}\:{root}\:{c}\:{and}\:{d} \\ $$$${then}\:{a},\:{b},\:{c}\:{and}\:{d}\:{are}\:{different}\:{real}\:{number}. \\ $$$${What}\:{the}\:{value}\:{of}\:{a}\:+\:{b}\:+\:{c}\:+\:{d}\:=\:? \\ $$ Commented by Syaka last updated…
Question Number 133928 by mnjuly1970 last updated on 25/Feb/21 Answered by mathmax by abdo last updated on 25/Feb/21 $$\left.\mathrm{2}\right)\:\mathrm{u}_{\mathrm{n}} =\left(\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \right)^{\frac{\mathrm{1}}{\mathrm{n}}} \:\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \:=\frac{\left(\mathrm{2n}\right)!}{\left(\mathrm{n}!\right)^{\mathrm{2}}…
Question Number 2857 by Syaka last updated on 29/Nov/15 $${By}\:{Induction}\:{Prove}\:{that}\:: \\ $$$$ \\ $$$$\mathrm{12}\mid\left({n}^{\mathrm{4}} \:−\:{n}^{\mathrm{2}} \right) \\ $$ Commented by Filup last updated on 29/Nov/15…
Question Number 133925 by bemath last updated on 25/Feb/21 $$\:\mathrm{Given}\:\mathrm{vector}\:\overset{\rightarrow} {{a}}\:=\:\hat {\mathrm{i}}−\mathrm{2}\hat {\mathrm{j}}+\hat {\mathrm{k}}\:,\: \\ $$$$\overset{\rightarrow} {{b}}=\:\mathrm{2}\hat {\mathrm{i}}+\hat {\mathrm{j}}−\mathrm{2}\hat {\mathrm{k}}\:,\:\overset{\rightarrow} {{c}}=−\hat {\mathrm{i}}+\mathrm{3}\hat {\mathrm{j}}−\hat {\mathrm{k}} \\…
Question Number 68390 by TawaTawa last updated on 10/Sep/19 Answered by som(math1967) last updated on 10/Sep/19 $${tan}\left(\mathrm{2tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{2}{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}}\:\right)+{tan}\theta \\ $$$$={tan}\left\{\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{cot}\:\frac{\theta}{\mathrm{2}}\right)\right\}+\mathrm{tan}\:\theta \\ $$$$={tan}\left\{\mathrm{2tan}^{−\mathrm{1}}…