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Question Number 100561 by DGmichael last updated on 27/Jun/20 Commented by bramlex last updated on 27/Jun/20 $${x}−{y}\:=\:\frac{{x}^{\mathrm{3}} −{y}^{\mathrm{3}} }{{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} } \\ $$$${x}=\sqrt[{\mathrm{3}\:}]{\mathrm{3}+\sqrt{\mathrm{9}+\frac{\mathrm{125}}{\mathrm{27}}}\:}\:,\:{y}\:=\:\sqrt[{\mathrm{3}\:}]{−\mathrm{3}+\sqrt{\mathrm{9}+\frac{\mathrm{125}}{\mathrm{27}}}} \\ $$$${x}−{y}\:=\:\frac{\mathrm{6}}{\:\sqrt[{\mathrm{3}\:}]{\left(\mathrm{3}+\sqrt{\mathrm{9}+\frac{\mathrm{125}}{\mathrm{27}}}\right)^{\mathrm{2}}…
Question Number 166089 by RB95 last updated on 13/Feb/22 Commented by RB95 last updated on 13/Feb/22 $$\mathrm{Help}\:\mathrm{me}\:\mathrm{please} \\ $$$$ \\ $$ Commented by nurtani last…
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Question Number 166077 by mkam last updated on 12/Feb/22 $${find}\:\int\:\frac{{cosx}}{\mathrm{1}+{cos}\left({x}\right)^{{tanx}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 166075 by abdurehime last updated on 12/Feb/22 $$\mathrm{find}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{and}\:\mathrm{range}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{relation}\:\left\{\left(\mathrm{x},\mathrm{y}\right):\mid\mathrm{x}\mid+\mathrm{y}\geq\mathrm{2}\right\}\:\mathrm{by}\:\mathrm{draw}\:\mathrm{its}\:\mathrm{graph} \\ $$ Commented by abdurehime last updated on 13/Feb/22 $$\mathrm{please}\:\mathrm{help}\:\mathrm{me} \\ $$ Terms…
Question Number 166049 by SANOGO last updated on 12/Feb/22 $${n}\epsilon\:{R}/\left\{\mathrm{0},\mathrm{1}\right\}\:{montrer}\:{que} \\ $$$$\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}\:\frac{{x}^{{k}} }{{nx}+{ln}\left({k}\right)}>=\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 166043 by mkam last updated on 12/Feb/22 $$\boldsymbol{{is}}\:\mathrm{811}\:\boldsymbol{{prime}}\:\boldsymbol{{number}}\:\boldsymbol{{or}}\:\boldsymbol{{no}}\:? \\ $$ Commented by mahdipoor last updated on 12/Feb/22 $${yes} \\ $$ Commented by mkam…
Question Number 166017 by naka3546 last updated on 11/Feb/22 $$\mathrm{Solve}\:\:\mathrm{it}\:! \\ $$$$\:\:\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\right)}\:−\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}−{t}\right)\:=\:\mathrm{tan}^{−\mathrm{1}} \:{t}\:−\:\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$ Terms of Service Privacy Policy Contact:…