Question Number 193821 by leandrosriv02 last updated on 20/Jun/23 Commented by talminator2856792 last updated on 21/Jun/23 $$\:\:−\left({a}\:+\:{b}\:−\:\mathrm{1}\right)\:+\:{a}\:+\:{b} \\ $$$$\:\:=\:{a}\:−{a}\:+\:{b}\:−{b}\:+\mathrm{1} \\ $$$$\:\:=\:\mathrm{0}\:+\:\mathrm{0}\:+\mathrm{1} \\ $$$$\:\:=\:\mathrm{1} \\ $$…
Question Number 193790 by pascal889 last updated on 20/Jun/23 Answered by Subhi last updated on 20/Jun/23 $$\:\mathrm{2}^{\mathrm{4}} \sqrt{{x}}\:+\frac{\mathrm{2}}{\:^{\mathrm{4}} \sqrt{{x}}}\:=\:\mathrm{5} \\ $$$$\sqrt{{x}}−\frac{\mathrm{5}}{\mathrm{2}}\:^{\mathrm{4}} \sqrt{{x}}\:+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:^{\mathrm{4}} \sqrt{{x}}\:=\:\frac{\frac{\mathrm{5}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{25}}{\mathrm{4}}−\mathrm{4}}}{\mathrm{2}}=\frac{\frac{\mathrm{5}}{\mathrm{2}}\pm\frac{\mathrm{3}}{\mathrm{2}}}{\mathrm{2}}=\:\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:\mathrm{2}…
Question Number 193835 by Skabetix last updated on 20/Jun/23 Commented by a.lgnaoui last updated on 21/Jun/23 $$\:\mathrm{respecter}\:\mathrm{les}\:\mathrm{mesures}\:\mathrm{pour}\:\mathrm{avoir}\:\mathrm{une} \\ $$$$\mathrm{idee}\:\mathrm{de}\:\mathrm{la}\:\mathrm{question} \\ $$$${Je}\:{crois}\:{que}\:{c}\:{est}\:{la}\:{premuere}\:{fois}\:{que} \\ $$$${tu}\:{envoie}\:{une}\:{question}\:{au}\:{forum}. \\ $$$$\mathrm{l}\:\mathrm{inconu}\:\mathrm{dans}\:\mathrm{votre}\:\mathrm{question}\:\mathrm{est}\:\boldsymbol{\mathrm{DE}}…
Question Number 193785 by aba last updated on 19/Jun/23 $${prove}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{5}}{\mathrm{6}}×\frac{\mathrm{7}}{\mathrm{8}}×\frac{\mathrm{9}}{\mathrm{10}}×…×\frac{\mathrm{99}}{\mathrm{100}}<\frac{\mathrm{1}}{\mathrm{10}}\:\:\: \\ $$ Answered by MM42 last updated on 20/Jun/23 $${a}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{5}}{\mathrm{6}}…×\frac{\mathrm{99}}{\mathrm{100}}< \\ $$$$<\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{6}}{\mathrm{7}}×…×\frac{\mathrm{98}}{\mathrm{99}}×\frac{\mathrm{100}}{\mathrm{101}}= \\…
Question Number 193750 by thean last updated on 19/Jun/23 Answered by Subhi last updated on 19/Jun/23 $${only}\:{one}\:{number}\:{can}\:{be}\:{zero}\:{to}\:{give}\:{the}\:{mini}\:{value} \\ $$$${suppose}\:{b}=\mathrm{0}\:\Rrightarrow\:{a}+{c}=\mathrm{2} \\ $$$$\frac{{a}}{\mathrm{3}{a}^{\mathrm{2}} }+\frac{{c}}{\mathrm{3}{c}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right) \\ $$$$\left({a}+{b}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\geqslant\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 193767 by sciencestudentW last updated on 19/Jun/23 $${prove}\:{that}\:{c}^{{log}_{{b}} {a}} ={a}^{{log}_{{b}} {c}} \\ $$ Answered by AST last updated on 19/Jun/23 $${c}^{{log}_{{b}} {a}} ={c}^{\frac{{log}\:{a}}{{log}\:{b}}=}…
Question Number 193721 by sciencestudentW last updated on 18/Jun/23 $${prove}\:{that}\:{a}^{{log}_{{a}} {N}} ={N} \\ $$ Answered by MATHEMATICSAM last updated on 18/Jun/23 $$\mathrm{From}\:\mathrm{log}\:\mathrm{definition}\:\mathrm{we}\:\mathrm{know}\:\mathrm{if} \\ $$$${a}^{{x}} \:=\:{N}\:\mathrm{then}\:\mathrm{we}\:\mathrm{can}\:\mathrm{write}\:\mathrm{it}\:\mathrm{as}…
Question Number 193715 by yaslm last updated on 18/Jun/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 193676 by lapache last updated on 18/Jun/23 $${Solve} \\ $$$${arcsin}\left(\mathrm{2}{x}\right)+{arcsin}\left({x}\sqrt{\mathrm{3}}\right)={arcsin}\left({x}\right) \\ $$ Answered by Frix last updated on 18/Jun/23 $$\mathrm{sin}^{−\mathrm{1}} \:{t}\:\in\mathbb{R}\:\Leftrightarrow\:−\mathrm{1}\leqslant{t}\leqslant\mathrm{1} \\ $$$${f}\left({x}\right)=\mathrm{sin}^{−\mathrm{1}}…
Question Number 193677 by lapache last updated on 18/Jun/23 $${li}\underset{{x}\rightarrow\mathrm{1}} {{m}}\:\frac{\mathrm{2}{arctan}^{\mathrm{2}} {x}−\frac{\pi}{\mathrm{8}}}{{x}^{\mathrm{2}} −\mathrm{1}}=… \\ $$ Answered by cortano12 last updated on 18/Jun/23 $$\:\:\underline{\underbrace{\:}} \cancel{\:} \\…