Question Number 99005 by bobhans last updated on 18/Jun/20 $$\underset{\mathrm{m}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{mn}\left(\mathrm{m}+\mathrm{n}\right)}\:?\: \\ $$ Answered by maths mind last updated on 18/Jun/20 $$\underset{{m}\geqslant\mathrm{1}}…
Question Number 164533 by alephzero last updated on 18/Jan/22 $$\mathrm{Prove},\:\mathrm{that} \\ $$$$\left.\mathrm{1}\right)\:\underset{{k}=\mathrm{1}} {\overset{\alpha} {\sum}}\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)}\:=\:\frac{\alpha}{\alpha+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}\left({k}+{x}\right)}\:=\:\mathrm{0} \\ $$ Answered by amin96 last…
Question Number 98983 by M±th+et+s last updated on 17/Jun/20 $${let}\:{a},{b},{c}\:{be}\:{positive}\:{real}\:{numbers}\:{such} \\ $$$${that}\:{ab}+{bc}+{ac}=\mathrm{3}\: \\ $$$${prove}\:{the}\:{inquality} \\ $$$$ \\ $$$$\frac{{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} +{bc}}+\frac{{b}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} +{ac}}+\frac{{c}\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}}…
Question Number 164511 by HongKing last updated on 18/Jan/22 $$\mathrm{sin}\centerdot\left(\pi\:\mathrm{sin}\:\mathrm{x}\right)\:-\:\mathrm{cos}\centerdot\left(\pi\:\mathrm{sin}\:\mathrm{x}\right)\:=\:\mathrm{1} \\ $$$$\mathrm{find}\:\:\boldsymbol{\mathrm{x}}=? \\ $$ Answered by mindispower last updated on 18/Jan/22 $${sin}\left({a}\right)−{cos}\left({a}\right)=\sqrt{\mathrm{2}}{sin}\left({a}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\Leftrightarrow\:{withea}=\pi{sin}\left({x}\right) \\…
Question Number 164496 by HongKing last updated on 18/Jan/22 $$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{xy}\:=\:\mathrm{11}}\\{\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}\:=\:\mathrm{24}}\end{cases}\:\:\Rightarrow\:\mid\mathrm{x}+\mathrm{y}\mid=? \\ $$ Answered by Rasheed.Sindhi last updated on 18/Jan/22 $$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{xy}\:=\:\mathrm{11}}\\{\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}\:=\:\mathrm{14}}\end{cases}\:\:\Rightarrow\:\mid\mathrm{x}+\mathrm{y}\mid=? \\…
Question Number 164497 by mnjuly1970 last updated on 18/Jan/22 Answered by cortano1 last updated on 18/Jan/22 Commented by mnjuly1970 last updated on 18/Jan/22 $${thanks}\:{alot}\:{sir}\:{cortano} \\…
Question Number 164466 by Zaynal last updated on 17/Jan/22 Commented by Zaynal last updated on 17/Jan/22 $$\left(\mathrm{xy}\right)^{\mathrm{2}} \:+\left({yz}\right)^{\mathrm{2}} +\left(\boldsymbol{{zx}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$ Terms of Service…
Question Number 98925 by ajfour last updated on 17/Jun/20 $${If}\:{the}\:{curve}\:{shown}\:{below}\:{has}\:{the}\: \\ $$$${equation},\:\:{y}=\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{bx}−{c}\right) \\ $$$${then}\:{find}\:\:{q}/{p}\:\:{in}\:{terms}\:{of}\:{b}\:{and}\:{c}. \\ $$ Commented by ajfour last updated on 17/Jun/20 Answered…
Question Number 164457 by mathls last updated on 17/Jan/22 $$\frac{\mathrm{20}}{\mathrm{10}}=\mathrm{2}\frac{\mathrm{0}}{\mathrm{10}}=>\frac{\mathrm{20}}{\mathrm{10}}=\mathrm{0}\:\:\:\:\:{why}? \\ $$ Answered by mahdipoor last updated on 17/Jan/22 $$\frac{\mathrm{20}}{\mathrm{10}}=\mathrm{2}\frac{\mathrm{0}}{\mathrm{10}}=\mathrm{2}+\frac{\mathrm{0}}{\mathrm{10}}\:\left({not}\:\mathrm{2}×\frac{\mathrm{0}}{\mathrm{10}}\right) \\ $$ Terms of Service…
Question Number 164451 by Zaynal last updated on 17/Jan/22 $$\boldsymbol{\mathrm{How}}\:\boldsymbol{\mathrm{do}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{true}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{false}}??; \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{Prove}}\:\boldsymbol{{the}}: \\ $$$$\:\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{\mathrm{p}}+\boldsymbol{\mathrm{p}}^{\mathrm{2}} +\boldsymbol{\mathrm{p}}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{q}}+\boldsymbol{{q}}^{\mathrm{2}} +\boldsymbol{{q}}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{r}}+\boldsymbol{{r}}^{\mathrm{2}} +\boldsymbol{{r}}^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{s}}+\boldsymbol{{s}}^{\mathrm{2}} +\boldsymbol{{s}}^{\mathrm{3}\:} }\:\geqslant\:\mathrm{1} \\…