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Category: Algebra

Question-12126

Question Number 12126 by tawa last updated on 13/Apr/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Apr/17 $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{ac}+\mathrm{2}{bc}=\mathrm{0} \\ $$$$\Rightarrow{ab}+{bc}+{ca}=−\mathrm{21} \\ $$$${A}=\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right) \\…

10-25-28-125-82-625-

Question Number 143184 by bramlexs22 last updated on 11/Jun/21 $$\:\:\:\:\frac{\mathrm{10}}{\mathrm{25}}+\frac{\mathrm{28}}{\mathrm{125}}+\frac{\mathrm{82}}{\mathrm{625}}+…\:=\:? \\ $$ Answered by Canebulok last updated on 11/Jun/21 $$\boldsymbol{{Solution}}: \\ $$$${In}\:{terms}\:{of}\:{summation}, \\ $$$$\Rightarrow\:\underset{{n}=\mathrm{1}} {\overset{\infty}…

Question-12080

Question Number 12080 by sin (x) last updated on 12/Apr/17 Answered by ajfour last updated on 12/Apr/17 $${f}\left(−\mathrm{1},\mathrm{3}\right)=\mathrm{max}\left(−\mathrm{4},−\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$${g}\left(\mathrm{4},\mathrm{4}\right)={min}\left(\mathrm{8},\:\mathrm{16}\right)\:=\mathrm{8} \\ $$$${f}\left[{f}\left(−\mathrm{1},\mathrm{3}\right),\:{g}\left(\mathrm{4},\mathrm{4}\right)\right] \\ $$$$\:\:\:=\:{f}\left(−\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{8}\right)={max}\left(−\frac{\mathrm{25}}{\mathrm{3}},\:−\frac{\mathrm{1}}{\mathrm{24}}\right) \\…

given-a-b-b-c-c-a-a-b-b-c-c-a-1-30-what-the-value-of-b-a-b-c-b-c-a-c-a-

Question Number 77604 by jagoll last updated on 08/Jan/20 $${given} \\ $$$$\frac{\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)}{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}=−\frac{\mathrm{1}}{\mathrm{30}} \\ $$$${what}\:{the}\:{value}\:{of}\: \\ $$$$\frac{{b}}{{a}+{b}}+\frac{{c}}{{b}+{c}}+\frac{{a}}{{c}+{a}}\:? \\ $$ Answered by key of knowledge last updated…

Let-a-b-gt-0-and-a-b-2-Prove-that-1-a-2-1-b-2-1-1-ab-2-

Question Number 143138 by loveineq last updated on 10/Jun/21 $$\mathrm{Let}\:{a},{b}\:>\:\mathrm{0}\:\mathrm{and}\:{a}+{b}\:=\:\mathrm{2}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}+{a}^{\mathrm{2}} \right)\left(\mathrm{1}+{b}^{\mathrm{2}} \right)\:\leqslant\:\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{ab}}}\right)^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…