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Category: Algebra

7-67-667-6667-n-terms-

Question Number 175411 by Rasheed.Sindhi last updated on 29/Aug/22 $$ \\ $$$$\mathrm{7}+\mathrm{67}+\mathrm{667}+\mathrm{6667}+…..\left({n}\:{terms}\right)=? \\ $$$$ \\ $$ Commented by infinityaction last updated on 30/Aug/22 $$\left(\mathrm{6}+\mathrm{1}\right)+\left(\mathrm{66}+\mathrm{1}\right)+\left(\mathrm{666}+\mathrm{1}\right)+…\left({n}\:{terms}\right) \\…

Question-175396

Question Number 175396 by Shrinava last updated on 29/Aug/22 Answered by mahdipoor last updated on 29/Aug/22 $${x}+{y}^{\mathrm{2}} \sqrt{\frac{\mathrm{2}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\geqslant{y}+\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}}\:\:\:\Leftrightarrow \\ $$$$\left({x}+{y}^{\mathrm{2}} \sqrt{\frac{\mathrm{2}}{{x}^{\mathrm{2}}…

if-x-1-4-4-x-1-find-x-

Question Number 175399 by Engr_Jidda last updated on 29/Aug/22 $${if}\:\left({x}+\mathrm{1}\right)^{\mathrm{4}} \sqrt{\mathrm{4}}={x}+\mathrm{1}\:\:{find}\:{x} \\ $$ Commented by Frix last updated on 29/Aug/22 $$\left({x}+\mathrm{1}\right)^{\mathrm{4}} ×\sqrt{\mathrm{4}}\:\mathrm{or}\:\left({x}+\mathrm{1}\right)×\sqrt[{\mathrm{4}}]{\mathrm{4}}? \\ $$ Commented…