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Category: Algebra

prove-for-real-x-y-and-a-that-x-a-2-y-2-x-a-2-y-2-2-x-2-y-2-

Question Number 13929 by ajfour last updated on 25/May/17 $${prove}\:{for}\:{real}\:\boldsymbol{{x}},\boldsymbol{{y}}\:{and}\:\boldsymbol{{a}}\:{that} \\ $$$$\sqrt{\left(\boldsymbol{{x}}+\boldsymbol{{a}}\right)^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} }+\sqrt{\left(\boldsymbol{{x}}−\boldsymbol{{a}}\right)^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} }\geqslant\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:. \\ $$$$ \\ $$ Answered by mrW1…

lcm-2a-3a-lcm-45-100-a-

Question Number 144989 by mathdanisur last updated on 01/Jul/21 $${lcm}\left(\mathrm{2}{a};\mathrm{3}{a}\right)={lcm}\left(\mathrm{45};\mathrm{100}\right)\Rightarrow{a}=? \\ $$ Answered by Rasheed.Sindhi last updated on 01/Jul/21 $${lcm}\left(\mathrm{2}{a};\mathrm{3}{a}\right)={lcm}\left(\mathrm{45};\mathrm{100}\right)\Rightarrow{a}=? \\ $$$${lcm}\left(\mathrm{2}{a};\mathrm{3}{a}\right)=\mathrm{6}{a}={lcm}\left(\mathrm{45};\mathrm{100}\right)=\mathrm{900} \\ $$$${a}=\mathrm{900}/\mathrm{6}=\mathrm{150} \\…

k-1-35-k-k-k-2-k-

Question Number 144981 by mathdanisur last updated on 01/Jul/21 $$\underset{{k}=\mathrm{1}} {\overset{\mathrm{35}} {\sum}}\:\frac{\sqrt{{k}}}{{k}\:+\:\sqrt{{k}^{\mathrm{2}} \:+\:{k}}}\:=\:? \\ $$ Answered by liberty last updated on 01/Jul/21 $$\:\frac{\sqrt{\mathrm{k}}}{\mathrm{k}+\sqrt{\mathrm{k}^{\mathrm{2}} +\mathrm{k}}}\:=\:\frac{\sqrt{\mathrm{k}}\:\left(\mathrm{k}−\sqrt{\mathrm{k}^{\mathrm{2}} +\mathrm{k}}\right)}{−\mathrm{k}}…

x-Z-15-48a-1-x-mod-17-find-x-

Question Number 144961 by mathdanisur last updated on 30/Jun/21 $${x}\in\mathbb{Z}^{+} \\ $$$$\mathrm{15}^{\mathrm{48}\boldsymbol{{a}}+\mathrm{1}} \:\equiv\:{x}\:\left({mod}\:\mathrm{17}\right)\:\:{find}\:\:{x}=?\: \\ $$ Answered by Rasheed.Sindhi last updated on 01/Jul/21 $$\mathrm{15}^{\mathrm{8}} \equiv{x}\left({mod}\mathrm{17}\right) \\…

Let-a-b-gt-0-and-a-b-1-3ab-Prove-that-1-a-2-a-1-b-2-b-1-a-a-2-1-b-b-2-1-2-a-2-b-1-b-2-a-1-a-b-2-1-b-a-2-1-

Question Number 144951 by loveineq last updated on 30/Jun/21 $$\mathrm{Let}\:{a},{b}\:>\:\mathrm{0}\:\mathrm{and}\:{a}+{b}+\mathrm{1}\:=\:\mathrm{3}{ab}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{2}} }{{a}+\mathrm{1}}+\frac{{b}^{\mathrm{2}} }{{b}+\mathrm{1}}\:\geqslant\:\frac{{a}}{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{{b}}{{b}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{2}} }{{b}+\mathrm{1}}+\frac{{b}^{\mathrm{2}} }{{a}+\mathrm{1}}\:\geqslant\:\frac{{a}}{{b}^{\mathrm{2}} +\mathrm{1}}+\frac{{b}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$ Commented…