Question Number 156689 by MathSh last updated on 14/Oct/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 25619 by aplus last updated on 12/Dec/17 Answered by prakash jain last updated on 12/Dec/17 $${ab}+{cd}=\mathrm{38}\:\:\left({i}\right) \\ $$$${ac}+{bd}=\mathrm{34}\:\:\:\left({ii}\right) \\ $$$${ad}+{bc}=\mathrm{43}\:\:\:\left({iii}\right) \\ $$$${add}\:\left({i}\right)\:{and}\:\left({ii}\right) \\…
Question Number 156684 by mathdave last updated on 14/Oct/21 $${solve}\:{the}\:{D}.{E}\: \\ $$$$\left[\mathrm{1}+{e}^{\frac{{x}}{{y}}} \right]{dx}+{e}^{\frac{{x}}{{y}}} \left[\mathrm{1}−\frac{{x}}{{y}}\right]{dy}=\mathrm{0} \\ $$$${any}\:{one}\:{can}\:{help}\:{pls} \\ $$ Answered by som(math1967) last updated on 14/Oct/21…
Question Number 156661 by MathSh last updated on 13/Oct/21 $$\mathrm{if}\:\:\mathrm{1}\leqslant\mathrm{x}\:\:;\:\:\mathrm{y}\leqslant\mathrm{2}\:\:;\:\:\mathrm{3}\leqslant\mathrm{z}\:\:;\:\:\mathrm{t}\leqslant\mathrm{4}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{1}}{\mathrm{xz}+\mathrm{yt}+\mathrm{3}}\:+\:\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}\:\leqslant\:\frac{\mathrm{3}+\mathrm{xz}+\mathrm{yt}}{\mathrm{9}+\mathrm{z}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }\:+\:\frac{\mathrm{11}}{\mathrm{12}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 25589 by behi.8.3.4.17@gmail.com last updated on 11/Dec/17 $$\boldsymbol{{possible}}\:\boldsymbol{{or}}\:\boldsymbol{{not}}\:? \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{a}}^{\boldsymbol{{a}}} +\boldsymbol{{b}}^{\boldsymbol{{b}}} >\:\boldsymbol{{a}}^{\boldsymbol{{b}}} +\boldsymbol{{b}}^{\boldsymbol{{a}}} \\ $$$$\boldsymbol{{with}}\:\boldsymbol{{below}}\:\boldsymbol{{conditions}}: \\ $$$$\left.\:\boldsymbol{{case}}\:\mathrm{1}\right)\:\:\boldsymbol{{a}}>\boldsymbol{{b}}>\mathrm{1} \\ $$$$\left.\boldsymbol{{case}}\:\mathrm{2}\right)\:\:\:\:\mathrm{0}<\boldsymbol{{b}}<\boldsymbol{{a}}<\mathrm{1} \\ $$ Commented by…
Question Number 156662 by MathSh last updated on 13/Oct/21 $$\boldsymbol{\mathrm{I}}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{2}\:\mathrm{ln}\:\mathrm{x}\:-\:\mathrm{x}\:+\:\frac{\mathrm{1}}{\mathrm{2}}}{\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} \right)\:\mathrm{ln}^{\mathrm{3}} \:\mathrm{x}}\:\mathrm{dx}\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 156652 by Tawa11 last updated on 13/Oct/21 $$\mathrm{Sirs},\:\mathrm{please}\:\mathrm{give}\:\mathrm{me}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solutions}\:\mathrm{to}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{ineqality}. \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:>\:\:\:\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:\geqslant\:\:\:\:\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:<\:\:\:\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:\leqslant\:\:\:\:\mathrm{0} \\ $$ Answered…
Question Number 156648 by MathSh last updated on 13/Oct/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{integers}: \\ $$$$\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:-\:\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{xy}\:-\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{0} \\ $$ Answered by Rasheed.Sindhi last updated on 14/Oct/21…
Question Number 156642 by MathSh last updated on 13/Oct/21 $$\mathrm{For}\:\:\mathrm{x}\in\left(\mathrm{0};\infty\right)\:-\:\mathbb{Z}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\left\{\mathrm{x}\right\}^{\mathrm{3}} }{\left[\mathrm{x}\right]}\:+\:\frac{\left[\mathrm{x}\right]^{\mathrm{3}} }{\left\{\mathrm{x}\right\}}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{8}}\:\left(\mathrm{x}^{\mathrm{2}} \:+\:\left[\mathrm{x}\right]^{\mathrm{2}} \:+\:\left\{\mathrm{x}\right\}^{\mathrm{2}} \right) \\ $$$$\left[\ast\right]-\mathrm{GIF}\:\:\mathrm{and}\:\:\left\{\mathrm{x}\right\}=\mathrm{x}-\left[\mathrm{x}\right] \\ $$ Commented by ghimisi last…
Question Number 156639 by ajfour last updated on 13/Oct/21 $$\mathrm{x}^{\mathrm{3}} =\mathrm{x}+\mathrm{c} \\ $$$$\mathrm{x}^{\mathrm{4}} =\mathrm{x}^{\mathrm{2}} +\mathrm{cx} \\ $$$$\mathrm{let}\:\:\mathrm{cx}=\mathrm{kx}^{\mathrm{4}} +\mathrm{hx}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{kx}^{\mathrm{3}} +\mathrm{hx}=\mathrm{c} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{1}+\mathrm{c}\left(\mathrm{kx}^{\mathrm{2}} +\mathrm{h}\right)…