Category: Algebra

Question Number 164039 by ArielVyny last updated on 13/Jan/22 $${consider}\:{f}\:{function}\:{Df}=\left[\mathrm{0},\mathrm{1}\right] \\ $$$${f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right)\:{c}\in\left[\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right]\:{show}\:{that}\:{f}\left({c}\right)={f}\left({c}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$ Answered by Ar Brandon last updated on 13/Jan/22 $${f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right) \\ $$$${f}\left({c}+\mathrm{0}\right)={f}\left({c}+\mathrm{1}\right)\:,\:\mathrm{since}\:{f}\:<\:\mathrm{1}-\mathrm{periodic}…

Question Number 164024 by mathlove last updated on 13/Jan/22 Answered by som(math1967) last updated on 13/Jan/22 $${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =−\left({x}−{y}\right) \\ $$$$\left({x}−{y}\right)\left({x}+{y}\right)+\left({x}−{y}\right)=\mathrm{0} \\ $$$$\left({x}−{y}\right)\left({x}+{y}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}+{y}=−\mathrm{1}\:\left[{x}\neq{y}\right]…

Question Number 164019 by ajfour last updated on 13/Jan/22 $$\:\:{x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$$\:\Rightarrow\:\:{x}^{\mathrm{2}} +\frac{{d}}{{x}^{\mathrm{2}} }+{ax}+\frac{{c}}{{x}}+{b}=\mathrm{0} \\ $$$${say}\:\:\:\frac{\mathrm{1}}{{x}}={t}\:\:\Rightarrow \\ $$$$\:\:{x}^{\mathrm{2}} +{ax}+{b}+{dt}^{\mathrm{2}} +{ct}=\mathrm{0} \\ $$$${let}\:\:\:\:{x}^{\mathrm{2}}…

Question Number 164001 by bekzodjumayev last updated on 12/Jan/22 Commented by bekzodjumayev last updated on 12/Jan/22 $${Help} \\ $$ Answered by mahdipoor last updated on…

Question Number 98448 by Quvonchbek last updated on 14/Jun/20 $$\:\:\:\:\:\:\:\mathrm{6}^{\mathrm{273}} +\mathrm{8}^{\mathrm{273}} \:\::\mathrm{49}\:\:\:\boldsymbol{{prove}}\:\:\boldsymbol{{the}}\:\:\boldsymbol{{divi}\mathrm{s}{ion}} \\ $$ Commented by Rasheed.Sindhi last updated on 14/Jun/20 $$\:\:\:\:\:\:\mathrm{49}\:\mid\:\mathrm{2}^{\mathrm{273}} \left(\mathrm{3}^{\mathrm{273}} +\mathrm{4}^{\mathrm{273}} \right)…