Question Number 54786 by behi83417@gmail.com last updated on 10/Feb/19 $${solve}\:{for}:{a},{b},{c},{d}\in\boldsymbol{{R}} \\ $$$$\:\:\:\:\:\:{a}^{\mathrm{2}} ={b}+\sqrt{{c}} \\ $$$$\:\:\:\:\:\:{b}^{\mathrm{2}} ={c}+\sqrt{{d}} \\ $$$$\:\:\:\:\:\:{c}^{\mathrm{2}} ={d}+\sqrt{{a}} \\ $$$$\:\:\:\:\:\:{d}^{\mathrm{2}} ={a}+\sqrt{{b}} \\ $$ Commented…
Question Number 54775 by MJS last updated on 10/Feb/19 $$\mathrm{found}\:\mathrm{something}\:\mathrm{interesting},\:\mathrm{it}\:\mathrm{was}\:\mathrm{published} \\ $$$$\mathrm{by}\:\mathrm{Tschirnhaus}\:\mathrm{in}\:\mathrm{1683} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{reduce} \\ $$$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{to} \\ $$$${y}^{\mathrm{3}} +{py}+{q}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{and}\:\mathrm{further}\:\mathrm{to}…
Question Number 185840 by Shrinava last updated on 28/Jan/23 Commented by mr W last updated on 29/Jan/23 $${here}\:{someone}\:{likes}\:{the}\:{question}\: \\ $$$${without}\:{to}\:{know}\:{what}\:{the}\:{question}\:{is}. \\ $$ Terms of Service…
Question Number 185842 by Shrinava last updated on 28/Jan/23 Commented by mr W last updated on 29/Jan/23 $${here}\:{someone}\:{likes}\:{the}\:{question}\: \\ $$$${without}\:{to}\:{know}\:{what}\:{the}\:{question}\:{is}. \\ $$ Terms of Service…
Question Number 185843 by Shrinava last updated on 28/Jan/23 Commented by mr W last updated on 29/Jan/23 $${here}\:{someone}\:{likes}\:{the}\:{question}\: \\ $$$${without}\:{to}\:{know}\:{what}\:{the}\:{question}\:{is}. \\ $$ Terms of Service…
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Question Number 185780 by mr W last updated on 27/Jan/23 $${find}\:{all}\:{solutions}\:{for} \\ $$$${k}!{m}!={n}! \\ $$$$\left({k},{m},{n}\in\mathbb{N}\right) \\ $$$$\left({m}\geqslant{k}>\mathrm{1}\right) \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 120240 by sorankarami0012 last updated on 30/Oct/20 $$\rightarrow \\ $$ Answered by mathmax by abdo last updated on 30/Oct/20 $$\curvearrowleft\frown\rightarrow\infty \\ $$ Terms…
Question Number 120229 by Algoritm last updated on 30/Oct/20 Answered by benjo_mathlover last updated on 30/Oct/20 $$\sqrt{\mathrm{36}\left({x}^{\mathrm{2}} +\mathrm{1}\right)−{x}}\:+\sqrt{{x}^{\mathrm{2}} +\mathrm{36}\left({x}+\mathrm{1}\right)}\:=\:{x}^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{36}\left(\left[{x}+\mathrm{1}\right]^{\mathrm{2}} −\mathrm{2}{x}\right)−{x}}\:+\sqrt{{x}^{\mathrm{2}} +\mathrm{36}\left({x}+\mathrm{1}\right)}\:=\:{x}^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{36}\left({x}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 120228 by mathace last updated on 30/Oct/20 $${Please}\:{solve}\:{using}\:{special}\:{function}\:{like}\:{LambertW} \\ $$$${function}\:{or}\:{any}\:{other}\:{function}\:\left({if}\:{possibe}\right) \\ $$$$\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{{x}} +\mathrm{17}^{{x}} =\mathrm{25}{x} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com