Question Number 18803 by daffa22 last updated on 30/Jul/17
Commented by daffa22 last updated on 30/Jul/17
$$\mathrm{help} \\ $$
Answered by behi.8.3.4.1.7@gmail.com last updated on 30/Jul/17
$${l}=\frac{\sqrt[{\mathrm{3}}]{−\mathrm{2}}−\mathrm{2}}{\:\sqrt{\mathrm{10}}−\sqrt{\mathrm{14}}}=\frac{\mathrm{2}+\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\:\sqrt{\mathrm{14}}−\sqrt{\mathrm{10}}}=\frac{\mathrm{3}.\mathrm{25}}{.\mathrm{58}}=\mathrm{5}.\mathrm{6} \\ $$$${non}\:{of}\:{the}\:{options}. \\ $$
Commented by daffa22 last updated on 30/Jul/17
$${maybe}\:{l}'{hospital}? \\ $$
Commented by behi.8.3.4.1.7@gmail.com last updated on 30/Jul/17
$${it}\:{is}\:{not}\:{nessesery}\:{to}\:{use}\:{l}'{hopital} \\ $$$${rule}. \\ $$
Commented by daffa22 last updated on 30/Jul/17
$${ok}\:{sir},{thank}\:{you},\:{I}\:{actually}\:{found}\:{that}\:{too}\:{but}\:{i}\:{want}\:{to}\:{make}\:{sure}\:{of}\:{it} \\ $$