Đáp án:
$\dfrac{128}{3}$
Giải thích các bước giải:
$($ $4^{30}$ $.$ $27^{10}$ $+$ $8^{20}$ $.$ $3^{29}$ $)$ $:$ $($ $6^{30}$ $.$ $2^{25}$ $)$
$→$ $\dfrac{4^{30} . 27^{10} + 8^{20} . 3^{29}}{6^{30} . 2^{25}}$
$→$ $\dfrac{(2^{2})^{30} . (3^{3})^{10} + (2^{3})^{20} . 3^{29}}{(2.3)^{30} . 2^{25}}$
$→$ $\dfrac{2^{60} . 3^{30} + 2^{60} . 3^{29}}{2^{30} . 3^{30} . 2^{25}}$
$→$ $\dfrac{2^{60} . 3^{30} + 2^{60} . 3^{29}}{2^{55} . 3^{30} }$
$→$ $\dfrac{2^{55} . (2^{5} . 3 + 2^{5})}{2^{55} . 3^{30} }$
$→$ $\dfrac{128}{3}$
$FbBinhne2k88$
