Question Number 78456 by arkanmath7@gmail.com last updated on 17/Jan/20 $${for}\:{a}>\mathrm{0}\:{and}\:{b}>{a}+\mathrm{2}\:,\:\:{verify}\:{the}\:{follwing}\: \\ $$$${claim}: \\ $$$$\:\:\:\sum_{{n}=\mathrm{1}} ^{\:\:\infty} \:{n}\:\frac{{a}\left({a}+\mathrm{1}\right)\left({a}+\mathrm{2}\right)…\left({a}+{n}−\mathrm{1}\right)}{{b}\left({b}+\mathrm{1}\right)\left({b}+\mathrm{2}\right)…\left({b}+{n}−\mathrm{1}\right)}\:=\frac{{a}\left({b}−\mathrm{1}\right)}{\left({b}−{a}−\mathrm{1}\right)\left({b}−{a}−\mathrm{2}\right)} \\ $$ Answered by mind is power last updated…
Question Number 78453 by mathocean1 last updated on 17/Jan/20 $$\mathrm{ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{points} \\ $$$$\mathrm{A}\left(−\mathrm{5};−\mathrm{5}\right)\:\mathrm{B}\left(−\mathrm{5};\mathrm{10}\right)\:\mathrm{C}\left(\mathrm{15};−\mathrm{5}\right). \\ $$$$\mathrm{the}\:\mathrm{cartesian}\:\mathrm{equtions}\:\mathrm{of}\:\left(\mathrm{AB}\right);\:\left(\mathrm{AC}\right) \\ $$$$\mathrm{and}\:\left(\mathrm{BC}\right)\:\mathrm{are}\:\mathrm{respectively} \\ $$$$\mathrm{x}=−\mathrm{5} \\ $$$$\mathrm{y}=−\mathrm{5} \\ $$$$\mathrm{x}+\mathrm{y}=\mathrm{5} \\ $$$$ \\…
Question Number 143987 by liberty last updated on 20/Jun/21 $${If}\:{f}\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{6}\right)+{f}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}\right)=\mathrm{2}{x} \\ $$$$\forall{x}\in{R}\:{th}\mathrm{e}{n}\:{f}\left(−\mathrm{3}\right)+{f}\left(\mathrm{9}\right)−\mathrm{5}{f}\left(\mathrm{1}\right)=? \\ $$ Answered by mitica last updated on 20/Jun/21 $${x}=\mathrm{1}\Rightarrow{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{1}\right)=\mathrm{2}\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1} \\…
Question Number 78449 by mathocean1 last updated on 17/Jan/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{definition}\:\mathrm{of}\: \\ $$$$\mathrm{f}\left({x}\right)=\frac{−{x}}{\mid{x}\mid−{x}} \\ $$ Commented by Rio Michael last updated on 18/Jan/20 $${f}\left({x}\right)\:=\:\frac{{x}}{{x}−\mid{x}\mid} \\ $$$${so}\:{x}\:<\:\mathrm{0}\:{so}\:{that}\:{x}−\mid{x}\mid\:{should}\:{be}\:…
Question Number 143980 by mathdanisur last updated on 20/Jun/21 Answered by mitica last updated on 20/Jun/21 $${p}={a}+{b}+{c};{q}={ab}+{bc}+{ac};{r}={abc}\Rightarrow{q}^{\mathrm{2}} \geqslant\mathrm{3}{pr} \\ $$$$\mathrm{3}\Sigma\frac{{a}}{{b}}−\Sigma\frac{\mathrm{3}{a}+{b}+{c}}{{b}+{c}}=\mathrm{3}\Sigma\left(\frac{{a}}{{b}}−\frac{{a}}{{b}+{c}}\right)−\mathrm{3}= \\ $$$$\mathrm{3}\Sigma\frac{{ac}}{{b}\left({b}+{c}\right)}−\mathrm{3}=\frac{\mathrm{3}}{{abc}}\Sigma\frac{\left({ac}\right)^{\mathrm{2}} }{{b}+{c}}−\mathrm{3}\geqslant \\ $$$$\frac{\mathrm{3}}{{r}}\centerdot\frac{{q}^{\mathrm{2}}…
Question Number 78447 by arkanmath7@gmail.com last updated on 17/Jan/20 $${prove}\:{that}\:{the}\:{seq}\:{a}_{{n}} \:=\:\frac{{ncos}\left(\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}\right)}{{n}+\mathrm{1}} \\ $$$${has}\:{convergent}\:{subsequence} \\ $$ Answered by mind is power last updated on 17/Jan/20…
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Question Number 12909 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/May/17 Answered by nume1114 last updated on 07/May/17 $$\:\:\:\:{a}^{\mathrm{2}} \left({a}+\mathrm{1}\right)+{b}^{\mathrm{2}} \left({b}+\mathrm{1}\right)+\mathrm{5}{ab} \\ $$$$={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{5}{ab}…
Question Number 12908 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/May/17 Answered by 433 last updated on 07/May/17 $$\int_{\sqrt{\mathrm{2}}} ^{\mathrm{2}} \left(\frac{{x}^{\mathrm{2}} }{{x}+\left[{x}+\mathrm{1}\right]}\right){dx}+\int_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \left(\frac{{x}^{\mathrm{2}} }{{x}+\left[{x}+\mathrm{1}\right]}\right){dx}= \\ $$$$\int_{\sqrt{\mathrm{2}}}…
Question Number 78443 by Maclaurin Stickker last updated on 17/Jan/20 Answered by mr W last updated on 18/Jan/20 Commented by mr W last updated on…