Question Number 224015 by hardmath last updated on 14/Aug/25 $$\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{2}=\mathrm{abc} \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\:+\:\sqrt{\mathrm{c}}\:\leqslant\:\frac{\mathrm{3}}{\mathrm{2}}\:\sqrt{\mathrm{abc}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 224025 by Simurdiera last updated on 14/Aug/25 $${Resuelve}\:{la}\:{ecuaci}\acute {{o}n}\:{diferencial} \\ $$$$\left[\mathrm{4}{x}^{\mathrm{3}} {y}\:−\:\frac{{e}^{{xy}} }{{x}}\:+\:{y}\:\mathrm{ln}\left({x}\right)\:+\:{x}\:\sqrt[{\mathrm{3}}]{{x}\:−\:\mathrm{4}}\right]{dx}\:+\:\left[{x}^{\mathrm{4}} −\:\frac{{e}^{{xy}} }{{y}}\:+\:{x}\:\mathrm{ln}\left({x}\right)\:−\:{x}\right]{dy} \\ $$$${Help}\:…. \\ $$ Terms of Service Privacy…
Question Number 224016 by hardmath last updated on 14/Aug/25 $$\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{2}=\mathrm{abc} \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}+\mathrm{a}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}+\mathrm{b}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}+\mathrm{c}}}\:\leqslant\:\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 224017 by hardmath last updated on 14/Aug/25 $$\mathrm{x},\mathrm{y},\mathrm{z}>\mathrm{0} \\ $$$$\mathrm{xy}+\mathrm{yz}+\mathrm{zx}+\mathrm{2xyz}=\mathrm{1} \\ $$$$\mathrm{prove}\:\mathrm{that}: \\ $$$$\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }\:\leqslant\:\frac{\mathrm{3}\:\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$ Terms of Service Privacy…
Question Number 224018 by hardmath last updated on 14/Aug/25 $$\mathrm{x}\:\neq\:\mathrm{y} \\ $$$$\lambda\:\geqslant\:\mathrm{1} \\ $$$$\begin{cases}{\mathrm{x}\:+\:\lambda^{\mathrm{2}} \:=\:\left(\mathrm{y}\:−\:\lambda\right)^{\mathrm{2}} }\\{\mathrm{y}\:+\:\lambda^{\mathrm{2}} \:=\:\left(\mathrm{x}\:−\:\lambda\right)^{\mathrm{2}} }\end{cases} \\ $$$$\mathrm{Find}:\:\:\:\left(\frac{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} }{\mathrm{4}\lambda^{\mathrm{2}} \:−\:\mathrm{1}}\right)^{\mathrm{2025}} =\:\:? \\…
Question Number 223995 by Rojarani last updated on 13/Aug/25 Answered by Frix last updated on 13/Aug/25 $$\mathrm{Let}\:{y}={px}\wedge{z}={qx},\:\mathrm{solve}\:\mathrm{for}\:{x}^{−\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${x}^{−\mathrm{2}} =\begin{cases}{\mathrm{7}{p}^{\mathrm{2}} +\mathrm{9}{pq}+\mathrm{3}{q}^{\mathrm{2}} +\mathrm{13}{p}+\mathrm{9}{q}+\mathrm{7}}\\{\left(\mathrm{7}{p}^{\mathrm{2}} +\mathrm{13}{pq}+\mathrm{7}{q}^{\mathrm{2}}…
Question Number 224003 by mr W last updated on 13/Aug/25 Commented by mr W last updated on 13/Aug/25 $${the}\:{areas}\:{of}\:{two}\:{equilaterals}\:{are} \\ $$$${known}.\:{find}\:{the}\:{area}\:{of}\:{the}\:{third} \\ $$$${triangle}. \\ $$…
Question Number 223978 by fantastic last updated on 12/Aug/25 $${Guys}\:{my}\:{exams}\:{are}\:{starting} \\ $$$${from}\:{today}.{Wish}\:{me}\:{luck}! \\ $$ Commented by som(math1967) last updated on 12/Aug/25 $$\boldsymbol{{Best}}\:\boldsymbol{{of}}\:\:\boldsymbol{{luck}}\: \\ $$ Commented…
Question Number 223988 by MirHasibulHossain last updated on 12/Aug/25 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{DE}\:\mathrm{using}\:\mathrm{the}\:\mathrm{method}\:\mathrm{of}\:\mathrm{Frobenius}\::\: \\ $$$$\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\mathrm{y}''−\mathrm{2xy}'+\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{y}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 223990 by RoseAli last updated on 12/Aug/25 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{5}} −\mathrm{32}\:}{{x}−\mathrm{1}} \\ $$ Commented by prathita last updated on 12/Aug/25 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}} \\ $$…