Question Number 12074 by sin (x) last updated on 11/Apr/17 Answered by ajfour last updated on 11/Apr/17 Commented by ajfour last updated on 11/Apr/17 $${M}\:{is}\:{the}\:{mid}-{point}\:{of}\:{CD}.…
Question Number 77604 by jagoll last updated on 08/Jan/20 $${given} \\ $$$$\frac{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)}{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}=−\frac{\mathrm{1}}{\mathrm{30}} \\ $$$${what}\:{the}\:{value}\:{of}\: \\ $$$$\frac{{b}}{{a}+{b}}+\frac{{c}}{{b}+{c}}+\frac{{a}}{{c}+{a}}\:? \\ $$ Answered by key of knowledge last updated…
Question Number 143138 by loveineq last updated on 10/Jun/21 $$\mathrm{Let}\:{a},{b}\:>\:\mathrm{0}\:\mathrm{and}\:{a}+{b}\:=\:\mathrm{2}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}+{a}^{\mathrm{2}} \right)\left(\mathrm{1}+{b}^{\mathrm{2}} \right)\:\leqslant\:\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{ab}}}\right)^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 143139 by loveineq last updated on 10/Jun/21 $$\mathrm{Let}\:{a},{b},{c}\:>\:\mathrm{0}\:\mathrm{and}\:{a}+{b}+{c}\:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}+{a}^{\mathrm{2}} \right)\left(\mathrm{1}+{b}^{\mathrm{2}} \right)\left(\mathrm{1}+{c}^{\mathrm{2}} \right)\:\leqslant\:\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{abc}}}\right)^{\mathrm{3}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 143130 by liberty last updated on 10/Jun/21 Commented by liberty last updated on 10/Jun/21 Answered by EDWIN88 last updated on 10/Jun/21 $$\mathrm{let}\:\mathrm{y}=\mathrm{px} \\…
Question Number 143122 by loveineq last updated on 10/Jun/21 $$\mathrm{Let}\:{a},{b}\:\in\left[\mathrm{0},\mathrm{1}\right]\:\mathrm{and}\:{a}+{b}\:\leqslant\:\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{a}}+\frac{\mathrm{1}}{\mathrm{1}+{b}}+\frac{\mathrm{1}}{\mathrm{2}}\:\leqslant\:\frac{\mathrm{2}}{{a}+{b}}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 143105 by mathdanisur last updated on 10/Jun/21 $$\frac{{cos}\left(\mathrm{3}{x}\right)}{{sin}\left(\mathrm{2}{x}\right)}\:=\:\mathrm{0} \\ $$ Commented by Dwaipayan Shikari last updated on 10/Jun/21 $${cos}\left(\mathrm{3}{x}\right)=\mathrm{0}\:\:\: \\ $$$$\mathrm{3}{x}=\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\Rightarrow{x}=\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{6}}\:\:\:\:{k}\in\mathbb{Z} \\ $$$${sin}\left(\mathrm{2}{x}\right)\neq\mathrm{0}…
Question Number 143101 by Huy last updated on 10/Jun/21 $$\mathrm{Find}\:\mathrm{x}:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{x}+\mathrm{1}} =\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{x}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 12030 by tawa last updated on 09/Apr/17 Answered by mrW1 last updated on 10/Apr/17 $${I}\:{think}\:{none}\:{of}\:{the}\:{answers}\:{is}\:{correct}. \\ $$$${A}.\:{Y}=\mathrm{4} \\ $$$$\mathrm{2}_{\mathrm{9}} ×\left(\mathrm{34}\right)_{\mathrm{9}} =\mathrm{2}×\mathrm{31}=\mathrm{62} \\ $$$$\mathrm{3}_{\mathrm{5}}…
Question Number 77556 by behi83417@gmail.com last updated on 07/Jan/20 $$\mathrm{1}.\:\:\:\mathrm{1}×\mathrm{1}!×\mathrm{2}!+\mathrm{2}×\mathrm{2}!×\mathrm{3}!+\mathrm{3}×\mathrm{3}!×\mathrm{4}!+….=? \\ $$$$\mathrm{2}.\:\:\:\:\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{3}}}+\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{3}+\sqrt{\mathrm{4}}}+\frac{\mathrm{3}+\sqrt{\mathrm{4}}}{\mathrm{4}+\sqrt{\mathrm{5}}}+……=? \\ $$$$\mathrm{3}.\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{4}}}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}}+\sqrt{\mathrm{4}}+\sqrt{\mathrm{5}}}+….=?\: \\ $$ Commented by mr W last updated on 07/Jan/20 $$\mathrm{1}.\:\:\rightarrow\infty,\:{clear}…