Question Number 189131 by Shrinava last updated on 12/Mar/23
Answered by cortano12 last updated on 12/Mar/23
$$\sqrt{\mathrm{19}+\mathrm{2}\sqrt{\mathrm{4}×\mathrm{15}}}\:=\mathrm{2}+\sqrt{\mathrm{15}} \\ $$
Answered by BaliramKumar last updated on 12/Mar/23
$$\mathrm{2}.\:{solution}:− \\ $$$$\:\:\:\: \\ $$$${x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{4}{y}^{\mathrm{2}} −\mathrm{16}{y}−\mathrm{7} \\ $$$$\underset{} {\underbrace{{x}^{\mathrm{2}} +\mathrm{2}\left({x}\right)\left(\mathrm{3}\right)+\mathrm{3}^{\mathrm{2}} }}−\underset{} {\underbrace{\mathrm{3}^{\mathrm{2}} −\left(\mathrm{2}{y}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}{y}\right)\left(\mathrm{4}\right)−\mathrm{4}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −\mathrm{7}}} \\ $$$$\left({x}+\mathrm{3}\right)^{\mathrm{2}} −\left(\mathrm{2}{y}+\mathrm{4}\right)^{\mathrm{2}} \\ $$$$\left({x}+\mathrm{2}{y}+\mathrm{7}\right)\left({x}−\mathrm{2}{y}−\mathrm{1}\right) \\ $$
Commented by Shrinava last updated on 13/Mar/23
$$\mathrm{thankyou}\:\mathrm{dearSer} \\ $$