Question

[phuongan nguyenthi]

\(\dfrac{4^0\cdot9^3-6^9\cdot1200}{8^4\cdot3^{12}+6^{11}}\)

Giúp mình nhé:]]]

Answer 1

[Nguyễn Ngọc Khánh Huyền]

\(\dfrac{4^0\cdot9^3-6^9\cdot1200}{8^4\cdot3^{12}+6^{11}}\)

= \(\dfrac{1\cdot\left(3^2\right)^3-\left(3\cdot2\right)^9\cdot\left(5^2\cdot2^4\cdot3\right)}{\left(2^3\right)^4\cdot3^{12}+6^{11}}\)

= \(\dfrac{3^6-3^9\cdot2^9\cdot5^2\cdot2^4\cdot3}{2^{12}\cdot3^{12}+\left(2\cdot3\right)^{11}}\)

= \(\dfrac{3^6-3^{10}\cdot2^{13}\cdot5^2}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

như lời đã nói thì bn làm tiếp>:)))