Câu 1.36 :
a. $\frac{3^{12} + 3^{15}}{1+ 3^{3}}$ = $\frac{3^{12} + 3^{3} .3^{12}}{1+ 3^{3}}$= $\frac{3^{12} (1 +3^{3} )}{1+ 3^{3}}$ = $3^{12}$
b. $ 2:\left ( \frac{1}{2} - \frac{2}{3}\right )^{2} + 0,125^{3}. 8^{3}- \left ( -12 \right )^{4}:6^{4}$= $ 2:\left ( \frac{3}{6} - \frac{4}{6}\right )^{2} + (0,125.8)^{3} - \left (12 \right )^{4}:6^{4}$ = 2:$(\frac{-1}{6})^{2}$ + 1 - $\left ( \frac{12}{6} \right )^{4}$ = 2: $\frac{1}{36}$ + 1 - $2^{4}$ = 72 + 1 -16 = 57