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Category: Algebra

Let-a-b-c-be-distinct-real-numbers-such-that-a-b-c-b-c-a-c-a-b-0-then-prove-that-a-b-c-2-b-c-a-2-c-a-b-2-0-

Question Number 217764 by ArshadS last updated on 20/Mar/25 $$ \\ $$$$\mathrm{Let}\:\mathrm{a},\:\mathrm{b},\:\mathrm{c}\:\mathrm{be}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that} \\ $$$$\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}=\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{{a}}{\left({b}−{c}\right)^{\mathrm{2}} }+\frac{{b}}{\left({c}−{a}\right)^{\mathrm{2}} }+\frac{{c}}{\left({a}−{b}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$ Answered by…

Question-217733

Question Number 217733 by Samuel12 last updated on 19/Mar/25 Answered by vnm last updated on 19/Mar/25 $${x}^{\frac{\mathrm{1}}{\mathrm{ln}\:\left({e}^{{x}} −\mathrm{1}\right)}} ={e}^{\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\left({e}^{{x}} −\mathrm{1}\right)}} ={e}^{\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:{x}+\mathrm{ln}\:\frac{{e}^{{x}} −\mathrm{1}}{{x}}}} = \\ $$$${e}^{\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{ln}\:{x}}\mathrm{ln}\:\frac{{e}^{{x}}…

f-x-f-y-f-x-y-xy-f-x-

Question Number 217660 by ArshadS last updated on 17/Mar/25 $$\:{f}\left({x}\right)\:+\:{f}\left({y}\right)={f}\left({x}+{y}\right)+{xy}\: \\ $$$${f}\left({x}\right)=? \\ $$ Answered by vnm last updated on 19/Mar/25 $${f}\left({x}\right)+{f}\left(\mathrm{0}\right)={f}\left({x}+\mathrm{0}\right)+{x}\centerdot\mathrm{0}\:\Rightarrow\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)+{f}\left(−{x}\right)={f}\left({x}−{x}\right)+{x}\left(−{x}\right)=−{x}^{\mathrm{2}} \\…