Question Number 165699 by lapache last updated on 06/Feb/22 $${li}\underset{{x}\rightarrow\mathrm{2}} {{m}}\frac{{x}−\sqrt{{x}+\mathrm{2}}}{\:\sqrt{\mathrm{4}{x}+\mathrm{1}}\:−\mathrm{3}}=….??? \\ $$ Commented by cortano1 last updated on 07/Feb/22 $$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{x}−\sqrt{\mathrm{x}+\mathrm{2}}}{\:\sqrt{\mathrm{4x}+\mathrm{1}}−\mathrm{3}}\:×\frac{\sqrt{\mathrm{4x}+\mathrm{1}}+\mathrm{3}}{\:\sqrt{\mathrm{4x}+\mathrm{1}}+\mathrm{3}}×\frac{\mathrm{x}+\sqrt{\mathrm{x}+\mathrm{2}}}{\mathrm{x}+\sqrt{\mathrm{x}+\mathrm{2}}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{4x}+\mathrm{1}}+\mathrm{3}}{\mathrm{x}+\sqrt{\mathrm{x}+\mathrm{2}}}\:×\underset{{x}\rightarrow\mathrm{2}}…
Question Number 100158 by Algoritm last updated on 25/Jun/20 Answered by MJS last updated on 25/Jun/20 $${u}=\sqrt{{x}+\mathrm{3}}\:\Leftrightarrow\:{x}={u}^{\mathrm{2}} −\mathrm{3}\wedge{u}\geqslant\mathrm{0} \\ $$$$\left.\left(\left.{u}+\mathrm{4}\right)\right)\sqrt{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{11}}−\mathrm{3}\right)=\mathrm{3}{u}^{\mathrm{2}} −\mathrm{20} \\ $$$$\left({u}+\mathrm{4}\right)\sqrt{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{11}}=\mathrm{3}{u}^{\mathrm{2}}…
Question Number 165698 by azizmathhacker last updated on 06/Feb/22 Answered by alephzero last updated on 06/Feb/22 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{e}^{\mathrm{ln}\:\frac{{x}+\mathrm{1}}{{x}}} \:=\:{e}^{\mathrm{ln}\:\mathrm{1}} \:=\:\mathrm{1} \\ $$ Commented by chrisbridge…
Question Number 165673 by mkam last updated on 06/Feb/22 Answered by mr W last updated on 06/Feb/22 $${to}\:{select}\:\mathrm{3}\:{from}\:\mathrm{10}\:{students}\:{there} \\ $$$${are}\:{totally}\:{C}_{\mathrm{3}} ^{\mathrm{10}} =\mathrm{120}\:{possibilties}. \\ $$$$\mathrm{0}\:{girl}:\:{C}_{\mathrm{3}} ^{\mathrm{6}}…
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Question Number 165669 by mkam last updated on 06/Feb/22 Commented by mkam last updated on 06/Feb/22 $${help}\:{me}\:{msr}\:{w} \\ $$ Answered by mr W last updated…
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Question Number 165651 by MathsFan last updated on 06/Feb/22 $${If}\:{the}\:{equation}\:\left(\mathrm{2}{x}−\mathrm{1}\right)−{p}\left({x}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{0}, \\ $$$${where}\:{p}\:{is}\:{constant},\:{has}\:{imaginary} \\ $$$${roots},\:{deduce}\:{that}\:\mathrm{2}{p}^{\mathrm{2}} +{p}−\mathrm{1}\geqslant\mathrm{0}. \\ $$ Commented by mr W last updated on…
Question Number 100106 by Tinku Tara last updated on 24/Jun/20 $$\mathrm{APK}\:\mathrm{with}\:\mathrm{the}\:\mathrm{option}\:\mathrm{to}\:\mathrm{review} \\ $$$$\mathrm{a}\:\mathrm{post}\:\mathrm{in}\:\mathrm{forum}\:\mathrm{is}\:\mathrm{available}. \\ $$$$\mathrm{Review}\:\mathrm{is}\:\mathrm{essentially}\:\mathrm{same}\:\mathrm{as}\:\mathrm{comment} \\ $$$$\mathrm{however}\:\mathrm{an}\:\mathrm{editable}\:\mathrm{original} \\ $$$$\mathrm{post}\:\mathrm{is}\:\mathrm{included}\:\mathrm{in}\:\mathrm{editor}. \\ $$$$\mathrm{While}\:\mathrm{review}\:\mathrm{post}\:\mathrm{with}\:\mathrm{images} \\ $$$$\mathrm{image}\:\mathrm{is}\:\mathrm{used}\:\mathrm{as}\:\mathrm{background}\:\mathrm{and} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathrm{shapes}\:\mathrm{like}\:\mathrm{ovals}…
Question Number 34567 by jordan last updated on 08/May/18 $${x}\begin{vmatrix}{\mathrm{2}}\\{}\end{vmatrix}−\mathrm{2}{x}−\mathrm{15}=\mathrm{0} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 08/May/18 $${x}^{\mathrm{2}} {is}\:{aleays}\:{positive}\:{hence}\:\mid{x}^{\mathrm{2}} \mid={x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{15}=\mathrm{0}…