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In-ABC-holds-cot-A-cot-B-3-cot-C-1-2-27-1-4-

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

Solve-x-1-x-2-x-1-x-2-2x-1-x-1-2x-1-x-1-

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-217881

Question Number 217881 by OmoloyeMichael last updated on 23/Mar/25 Answered by vnm last updated on 23/Mar/25 $$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {x}\left(\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} {x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {x}}\right){dx}={I} \\ $$$${f}\left({x}\right)=\frac{\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} {x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {x}}…

Solve-2x-7-x-1-2-

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} \\…