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Category: Algebra

2-3-x-2-3-x-4-x-

Question Number 179529 by mathlove last updated on 30/Oct/22 $$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{x}} +\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{x}} =\mathrm{4}\:\:\:\:\:\:{x}=? \\ $$ Answered by mr W last updated on 30/Oct/22 $$\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)=\mathrm{2}^{\mathrm{2}} −\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{4}−\mathrm{3}=\mathrm{1}…

a-1-b-2cos-2pi-13-b-1-c-2cos-6pi-13-c-1-a-2cos-8pi-13-prove-that-abc-5-1-abc-5-11-

Question Number 179518 by mathlove last updated on 30/Oct/22 $${a}−\frac{\mathrm{1}}{{b}}=\mathrm{2}{cos}\frac{\mathrm{2}\pi}{\mathrm{13}} \\ $$$${b}−\frac{\mathrm{1}}{{c}}=\mathrm{2}{cos}\frac{\mathrm{6}\pi}{\mathrm{13}} \\ $$$${c}−\frac{\mathrm{1}}{{a}}=\mathrm{2}{cos}\frac{\mathrm{8}\pi}{\mathrm{13}}\:\:\:\:\:\:{prove}\:{that} \\ $$$$\left({abc}\right)^{\mathrm{5}} −\frac{\mathrm{1}}{\left({abc}\right)^{\mathrm{5}} }=\mathrm{11} \\ $$ Commented by mr W last…

Find-the-nth-term-1-0-1-0-1-0-1-0-

Question Number 113975 by I want to learn more last updated on 16/Sep/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{term}:\:\:\:\mathrm{1},\:\:\mathrm{0},\:\:−\:\mathrm{1},\:\:\mathrm{0},\:\:\mathrm{1},\:\:\mathrm{0},\:\:−\:\mathrm{1},\:\:\mathrm{0},\:\:… \\ $$ Answered by Olaf last updated on 16/Sep/20 $${u}_{{n}} \:=\:\mathrm{cos}\left({n}\frac{\pi}{\mathrm{2}}\right)…

Evaluer-1-logx-1-x-dx-

Question Number 179494 by a.lgnaoui last updated on 29/Oct/22 $${Evaluer} \\ $$$$\int\frac{\mathrm{1}−\mathrm{logx}}{\mathrm{1}+\mathrm{x}}\mathrm{dx} \\ $$ Answered by ARUNG_Brandon_MBU last updated on 29/Oct/22 $$\int\frac{\mathrm{1}−\mathrm{log}{x}}{\mathrm{1}+{x}}{dx} \\ $$$$=\mathrm{log}\left(\mathrm{1}+{x}\right)−\underset{{n}=\mathrm{0}} {\overset{\infty}…

x-2-x-2-x-2-OR-x-2-x-2-x-2-

Question Number 113931 by Rasheed.Sindhi last updated on 16/Sep/20 $$\mid\:{x}\:\mid=\mathrm{2}\Rightarrow\:{x}=\mathrm{2}\:\vee\:{x}=−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{OR} \\ $$$$\mid\:{x}\:\mid=\mathrm{2}\Rightarrow\:{x}=\mathrm{2}\:\wedge\:{x}=−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:? \\ $$ Commented by bemath last updated on 16/Sep/20…

Find-the-square-root-of-50-48-

Question Number 113913 by Aina Samuel Temidayo last updated on 16/Sep/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{square}\:\mathrm{root}\:\mathrm{of}\:\sqrt{\mathrm{50}}+\sqrt{\mathrm{48}} \\ $$ Answered by 1549442205PVT last updated on 16/Sep/20 $$\mathrm{We}\:\mathrm{have}\:\left(\sqrt{\mathrm{50}}+\sqrt{\mathrm{48}}\right)^{\mathrm{2}} =\mathrm{98}+\mathrm{2}\sqrt{\mathrm{50}.\mathrm{48}} \\ $$$$=\mathrm{98}+\mathrm{2}\sqrt{\mathrm{25}.\mathrm{16}.\mathrm{6}}=\mathrm{98}+\mathrm{40}\sqrt{\mathrm{6}}…