Question Number 151351 by mathdanisur last updated on 20/Aug/21 Answered by dumitrel last updated on 20/Aug/21 $$\:\lambda\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}}}+\frac{\mathrm{2}{ab}}{{a}+{b}}\geqslant\frac{\lambda+\mathrm{1}}{\mathrm{2}}\left({a}+{b}\right) \\ $$$${if}\:{a}={b}\:{true} \\ $$$${suppose}\:{a}<{b}\:;\frac{{a}}{{b}}={t}\in\left(\mathrm{0};\mathrm{1}\right) \\ $$$$\Leftrightarrow\lambda\sqrt{\frac{{t}^{\mathrm{2}}…
Question Number 151346 by mathdanisur last updated on 20/Aug/21 $$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{5x}\:+\:\mathrm{25}\:=\:\mathrm{0}\:\:\Rightarrow\:\:\mathrm{x}^{\mathrm{3}} \:-\:\mathrm{100}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 20/Aug/21 $$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{5x}\:+\:\mathrm{25}\:=\:\mathrm{0}\:\:\Rightarrow\:\:\mathrm{x}^{\mathrm{3}} \:-\:\mathrm{100}\:=\:? \\…
Question Number 151344 by mathdanisur last updated on 20/Aug/21 $$\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{4x}\:+\:\mathrm{1}\:=\:\mathrm{0}\:\:\Rightarrow\:\:\mathrm{x}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\:=\:? \\ $$ Answered by dumitrel last updated on 20/Aug/21 $${x}−\mathrm{4}+\frac{\mathrm{1}}{{x}}=\mathrm{0}\Rightarrow{x}+\frac{\mathrm{1}}{{x}}=\mathrm{4}\Rightarrow \\ $$$$\mathrm{64}={x}^{\mathrm{3}}…
Question Number 151332 by mathdanisur last updated on 20/Aug/21 $$\mathrm{which}\:\mathrm{is}\:\mathrm{larger}? \\ $$$$\mathrm{2}^{\sqrt{\mathrm{10}}} \:\:\mathrm{or}\:\:\mathrm{3}^{\mathrm{2}} \\ $$ Commented by liberty last updated on 20/Aug/21 $$\mathrm{3}^{\mathrm{2}} \\ $$…
Question Number 20259 by Tinkutara last updated on 24/Aug/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:+\:\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$ Answered by ajfour last updated on 24/Aug/17 $${f}\:'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\:>\mathrm{0}…
Question Number 151322 by mathdanisur last updated on 20/Aug/21 $$\mathrm{Determine}\:\mathrm{all}\:\mathrm{pairs}\:\:\left(\mathrm{x};\mathrm{y}\right)\:\:\mathrm{of}\:\mathrm{integers} \\ $$$$\mathrm{which}\:\mathrm{satisfy}: \\ $$$$\mathrm{x}^{\mathrm{3}} \:-\:\mathrm{3x}\:+\:\mathrm{y}\:=\:\mathrm{0} \\ $$ Commented by mathdanisur last updated on 20/Aug/21 $$\mathrm{with}\:\boldsymbol{\mathrm{x}}\:\mathrm{real},\:\boldsymbol{\mathrm{y}}\:\mathrm{integer}…
Question Number 151316 by iloveisrael last updated on 20/Aug/21 $$\:\:\:\:\:\:{Find}\:{positive}\:{integers}\: \\ $$$$\:\:\:\:\:{a}\:{and}\:{b}\:{such}\:{that}\: \\ $$$$\:\:\:\:\:\left(\sqrt[{\mathrm{3}}]{{a}}\:+\sqrt[{\mathrm{3}}]{{b}}\:−\mathrm{1}\right)^{\mathrm{2}} =\:\mathrm{49}+\:\mathrm{20}\sqrt[{\mathrm{3}}]{\mathrm{6}}\: \\ $$ Answered by Rasheed.Sindhi last updated on 20/Aug/21 $$\overset{\mathcal{SOLVE}\:\:{for}\:{a},{b}\in\mathbb{Z}^{+}…
Question Number 151317 by mathdanisur last updated on 20/Aug/21 $$\mathrm{if}\:\:\boldsymbol{\mathrm{x}}=\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{7}}\:-\:\sqrt{\mathrm{6}}}\:+\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{7}}\:+\:\sqrt{\mathrm{6}}} \\ $$$$\mathrm{find}\:\:\mathrm{E}\left(\mathrm{x}\right)\:=\:\mathrm{x}^{\mathrm{6}} \:-\:\mathrm{6x}^{\mathrm{4}} \:+\:\mathrm{9x}^{\mathrm{2}} \:=\:? \\ $$ Answered by iloveisrael last updated on 20/Aug/21 $$\:{we}\:{have}\:{x}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{7}}−\sqrt{\mathrm{6}}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{7}}+\sqrt{\mathrm{6}}}\:=\:\mathrm{0}…
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Question Number 151315 by mathdanisur last updated on 20/Aug/21 $$\mathrm{if}\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{16}^{\mathrm{2}} \:\:;\:\:\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{24}^{\mathrm{2}} \\ $$$$\mathrm{z}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} =\mathrm{42}^{\mathrm{2}} \:\:\mathrm{and}\:\:\mathrm{t}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} =\mathrm{38}^{\mathrm{2}} \\ $$$$\mathrm{find}\:\:\mathrm{max}\left[\left(\mathrm{x}+\mathrm{z}\right)\left(\mathrm{y}+\mathrm{t}\right)\right]=? \\…