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 $$\mathrm{1}.\:\:\rightarrow\infty,\:{clear}…
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Question Number 77554 by TawaTawa last updated on 07/Jan/20 Commented by TawaTawa last updated on 07/Jan/20 $$\mathrm{Please}\:\mathrm{help}\:\mathrm{with}\:\:\:\mathrm{1},\:\mathrm{2},\:\mathrm{3} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 12019 by Nayon last updated on 09/Apr/17 Answered by ajfour last updated on 10/Apr/17 Commented by ajfour last updated on 10/Apr/17 $${AC}^{\:\mathrm{2}} +{BD}^{\mathrm{2}}…
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Question Number 77552 by lémùst last updated on 07/Jan/20 $$\int\frac{\mathrm{x}−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}\:=\:? \\ $$ Answered by MJS last updated on 07/Jan/20 $$\int\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}−\mathrm{4}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}}…
Question Number 12017 by Nayon last updated on 09/Apr/17 $${How}\:{Can}\:{we}\:{expand}\:\left({a}+{b}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{and} \\ $$$$\left({a}+{b}\right)^{−{n}} \:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 12015 by Nayon last updated on 09/Apr/17 $${prove}\:{that}\:{for}\:{all}\:{x}\:\in{R}, \\ $$$${e}^{{x}} \geqslant{x}^{{e}} \\ $$ Answered by mrW1 last updated on 18/Apr/17 $${at}\:{x}={e},\:{e}^{{x}} ={x}^{{e}} \\…