Question Number 218019 by dscm last updated on 26/Mar/25 $${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{x}+{y}=\mathrm{18} \\ $$$${xy}+{x}+{y}=\mathrm{7} \\ $$$${Solve}. \\ $$ Answered by Ghisom last updated on 26/Mar/25…
Question Number 218003 by hardmath last updated on 25/Mar/25 $$\mathrm{1}.\:\:\:\int\:\:\frac{\mathrm{x}\:\mathrm{dx}}{\left(\mathrm{x}\:+\:\mathrm{2}\right)\centerdot\left(\mathrm{x}\:+\:\mathrm{4}\right)} \\ $$$$\mathrm{2}.\:\:\:\int\:\:\frac{\mathrm{x}\:+\:\mathrm{2}}{\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{2x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\mathrm{3}.\:\:\:\int\:\:\frac{\mathrm{2x}\:+\:\mathrm{7}}{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{4x}\:+\:\mathrm{3}}\:\mathrm{dx} \\ $$ Commented by Ghisom last updated on…
Question Number 217985 by hardmath last updated on 24/Mar/25 $$\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{3x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{The}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\rightarrow\:\mathrm{a}\:,\:\mathrm{b}\:,\:\mathrm{c} \\ $$$$\mathrm{Find}\:\rightarrow\:\sqrt[{\mathrm{3}}]{\mathrm{a}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{b}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{c}}\:=\:? \\ $$ Commented by Frix last updated on 24/Mar/25 $$\mathrm{suppose}\:{a}<{b}<{c}…
Question Number 217911 by Rasheed.Sindhi last updated on 23/Mar/25 $${Solve} \\ $$$$\left(\frac{{x}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{x}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} =\mathrm{6} \\ $$ Answered by Rasheed.Sindhi last updated on 23/Mar/25 $$\left(\frac{{x}}{{x}−\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{{x}}{{x}+\mathrm{1}}\right)^{\mathrm{2}}…
Question Number 217894 by hardmath last updated on 23/Mar/25 $$\mathrm{In}\:\:\bigtriangleup\mathrm{ABC}\:\:\mathrm{holds}: \\ $$$$\Sigma\:\frac{\mathrm{cot}\:\mathrm{A}}{\:\sqrt{\mathrm{cot}\:\mathrm{B}\:\:+\:\:\mathrm{3}\:\mathrm{cot}\:\mathrm{C}}}\:\:\geqslant\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\sqrt[{\mathrm{4}}]{\mathrm{27}}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 217895 by hardmath last updated on 23/Mar/25 Commented by mr W last updated on 23/Mar/25 $${question}\:{is}\:{wrong}.\:{C}_{{n}} ^{{r}} \:{is}\:{only}\:{defined} \\ $$$${for}\:\mathrm{0}\leqslant{r}\leqslant{n}. \\ $$ Commented…
Question Number 217888 by ArshadS last updated on 23/Mar/25 $${Solve}: \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{15}}=\frac{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{24}} \\ $$ Answered by bagjahebat last updated on 23/Mar/25…
Question Number 217887 by ArshadS last updated on 23/Mar/25 $${Solve}\:: \\ $$$$\frac{{x}+\mathrm{1}}{{x}−\mathrm{2}}+\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}=\frac{\mathrm{2}{x}+\mathrm{1}}{{x}−\mathrm{1}}+\frac{\mathrm{2}{x}−\mathrm{1}}{{x}+\mathrm{1}} \\ $$ Answered by Rasheed.Sindhi last updated on 23/Mar/25 $$\frac{{x}+\mathrm{1}}{{x}−\mathrm{2}}−\mathrm{1}+\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}−\mathrm{1}+\mathrm{2}=\frac{\mathrm{2}{x}+\mathrm{1}}{{x}−\mathrm{1}}−\mathrm{2}+\frac{\mathrm{2}{x}−\mathrm{1}}{{x}+\mathrm{1}}−\mathrm{2}+\mathrm{4} \\ $$$$\frac{\mathrm{3}}{{x}−\mathrm{2}}+\frac{−\mathrm{3}}{{x}+\mathrm{2}}=\frac{\mathrm{3}}{{x}−\mathrm{1}}+\frac{−\mathrm{3}}{{x}+\mathrm{1}}+\mathrm{2} \\…
Question Number 217949 by hardmath last updated on 23/Mar/25 Commented by hardmath last updated on 23/Mar/25 $$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}} \:=\:? \\ $$ Answered by A5T…
Question Number 217940 by ArshadS last updated on 23/Mar/25 $${Solve} \\ $$$$\sqrt{\mathrm{2}{x}+\mathrm{7}}\:\:−\sqrt{{x}−\mathrm{1}}\:=\mathrm{2}\:\:\:\:\:\:\:\:\:\:\: \\ $$ Answered by Frix last updated on 23/Mar/25 $$\mathrm{No}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{because}\:\mathrm{min}\:\left(\mathrm{lhs}\right)\:>\mathrm{2} \\ $$$${a},\:{b}\:\in\mathbb{R}\wedge{b}\neq\mathrm{0} \\…