Question

Calculati: $(a+ \frac{a^{2} }{3})-( \frac{a}{3}+2a^{2})+( \frac{5a}{6}- \frac{ a^{2} }{4} )$

Answer 1

$\frac{12a}{12}$+$\frac{4 a^{2} }{12}$-$\frac{4a}{12}$-$\frac{24 a^{2} }{12}$+$\frac{10a}{12}$-$\frac{3 a^{2} }{12}$=$\frac{18a- 23a^{2} }{12}$

Answer 2

$(a+ \frac{ a^{2} }{3})-( \frac{a}{3}+2 a^{2})+ (\frac{5a}{6}- \frac{ a^{2} }{4})= \\ =a+ \frac{ a^{2} }{3}- \frac{a}{3}- 2a^{2} + \frac{5a}{6}- \frac{ a^{2} }{4}= \\ =(a- \frac{a}{3}+ \frac{5a}{6})+( \frac{ a^{2} }{3}- 2a^{2}- \frac{ a^{2} }{4})= \\ =a(1- \frac{1}{3}+ \frac{5}{6})+ a^{2}( \frac{1}{3}-2- \frac{1}{4})= \\ =a( \frac{6}{6}- \frac{2}{6}+ \frac{5}{6})+ a^{2}( \frac{4}{12}- \frac{24}{12}- \frac{3}{12})= \\ =a* \frac{9}{6}+ a^{2} *(- \frac{23}{12})=$
$= \frac{18a}{12}- \frac{23 a^{2} }{12}= \\ = \frac{18a-23 a^{2} }{12}$