Question Number 165829 by HongKing last updated on 09/Feb/22 $$\mathrm{Compare}\:\mathrm{it}: \\ $$$$\mathrm{p}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:…\:+\:\frac{\mathrm{1}}{\mathrm{100}^{\mathrm{2}} }\:\:\mathrm{and}\:\:\mathrm{q}\:=\:\mathrm{0},\mathrm{99} \\ $$$$\left(\mathrm{a}\right)\mathrm{p}=\mathrm{q}\:\:\left(\mathrm{b}\right)\mathrm{p}<\mathrm{q}\:\:\left(\mathrm{c}\right)\mathrm{p}>\mathrm{q}\:\:\left(\mathrm{d}\right)\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{e}\right)\:\sqrt{\mathrm{q}}\:=\:\sqrt{\mathrm{p}}\:-\:\mathrm{2} \\ $$ Answered by…
Question Number 165830 by HongKing last updated on 09/Feb/22 $$\sqrt{\mathrm{8a}\:+\:\sqrt{\mathrm{8a}\:+\:\sqrt{\mathrm{8a}\:+\:…}}}\:-\:\sqrt{\mathrm{a}\:\sqrt{\mathrm{a}\:\sqrt{\mathrm{a}\:…}}}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{9}\:\:\left(\mathrm{b}\right)\:\mathrm{3}\:\:\left(\mathrm{c}\right)\:\mathrm{6}\:\:\left(\mathrm{d}\right)\:\mathrm{1}\:\:\left(\mathrm{e}\right)\mathrm{12} \\ $$ Answered by Rasheed.Sindhi last updated on 09/Feb/22 $$\sqrt{\mathrm{8a}\:+\:\sqrt{\mathrm{8a}\:+\:\sqrt{\mathrm{8a}\:+\:…}}}\:-\:\sqrt{\mathrm{a}\:\sqrt{\mathrm{a}\:\sqrt{\mathrm{a}\:…}}}\:=\:\mathrm{0} \\ $$$$\sqrt{\mathrm{8a}\:+\:\sqrt{\mathrm{8a}\:+\:\sqrt{\mathrm{8a}\:+\:…}}}\:=\:\sqrt{\mathrm{a}\:\sqrt{\mathrm{a}\:\sqrt{\mathrm{a}\:…}}}\:=\:\mathrm{b}\left(\mathrm{say}\right) \\…
Question Number 165831 by HongKing last updated on 09/Feb/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{integer}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{number}: \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{2015}\:\centerdot\:\mathrm{2016}\:\centerdot\:\mathrm{2017}} \\ $$ Commented by mr W last updated on 09/Feb/22 $$\sqrt[{\mathrm{3}}]{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)} \\ $$$$>\sqrt[{\mathrm{3}}]{{x}×{x}×{x}}={x}…
Question Number 34739 by chakraborty ankit last updated on 10/May/18 Commented by abdo mathsup 649 cc last updated on 10/May/18 $$\left.\mathrm{5}\right)\:{we}\:{have}\:{the}\:{formula} \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left[{x}\:+\frac{{k}}{{n}}\right]\:=\left[{nx}\right]\:\Rightarrow…
Question Number 34738 by chakraborty ankit last updated on 10/May/18 Commented by candre last updated on 10/May/18 $$\left\{\mathrm{1},\mathrm{2},\mathrm{3}\right\} \\ $$$$\left\{\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{2},\mathrm{2}\right),\left(\mathrm{3},\mathrm{3}\right),\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{2},\mathrm{1}\right),\left(\mathrm{2},\mathrm{3}\right),\left(\mathrm{3},\mathrm{2}\right),\left(\mathrm{1},\mathrm{3}\right),\left(\mathrm{3},\mathrm{1}\right)\right\} \\ $$$$\left({B}\right)\mathrm{7} \\ $$ Commented…
Question Number 165804 by cortano1 last updated on 08/Feb/22 Answered by MJS_new last updated on 09/Feb/22 $$\left(\mathrm{1}\right)\:{y}=−\frac{{x}\left(\mathrm{5}{x}−\mathrm{1}\right)}{\mathrm{3}\left({x}−\mathrm{1}\right)} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{2}\right)\:{x}^{\mathrm{3}} −{x}^{\mathrm{2}} =\frac{{x}^{\mathrm{2}} \left(\mathrm{5}{x}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{5}{x}^{\mathrm{2}}…
Question Number 165795 by mnjuly1970 last updated on 08/Feb/22 Answered by ArielVyny last updated on 08/Feb/22 $${soit}\:{f}\left({x},{y}\right)={x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+{y}^{\mathrm{2}} +\frac{{y}}{{x}} \\ $$$${montrons}\:{que}\:{f}\left({x},{y}\right)\geqslant\sqrt{\mathrm{3}}\:\:{x}\neq\mathrm{0} \\ $$$$-{fixons}\:{x}\in\mathbb{R}\:{on}\:{a}\:{f}_{{x}} \left({y}\right)={x}^{\mathrm{2}}…
Question Number 165774 by mathlove last updated on 08/Feb/22 Answered by Eulerian last updated on 08/Feb/22 $$\:\mathrm{let}:\:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{3}}\:=\:\mathrm{y} \\ $$$$\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{y}}+\mathrm{3}\:\:\Rightarrow\:\:\mathrm{y}\:=\:\frac{\mathrm{1}}{\mathrm{x}}\:+\:\mathrm{3} \\ $$$$\: \\ $$$$\:\therefore \\ $$$$\:\mathrm{f}\left(\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{x}}+\mathrm{3}−\mathrm{3}}\right)\:=\:\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{x}}\:+\:\mathrm{3}\right)}…
Question Number 34685 by math khazana by abdo last updated on 09/May/18 $${decompose}\:{the}\:{fraction} \\ $$$${F}\left({x}\right)=\:\:\frac{\mathrm{1}}{\left({x}+\mathrm{2}\right)\left(\:{x}^{{n}} \:\:−\mathrm{1}\right)}\:\:{with}\:{n}\:\in\:{N}^{\bigstar} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 34686 by math khazana by abdo last updated on 09/May/18 $${decompose}\:{F}\left({x}\right)\:\:=\:\frac{\left(\mathrm{2}{n}\right)!}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{2}} \:−\mathrm{2}\right)….\left({x}^{\mathrm{2}} \:−{n}\right)} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com