Question

\(\dfrac{27^4.4^3}{9^5.8^2}\);\(\dfrac{3^{29}.4^{16}}{27^9.8^{11}}\)

Answer 1

a)\(\dfrac{27^4.4^3}{9^5.8^2}\)

=\(\dfrac{3^{12}.2^6}{3^{10}.2^6}\)

=3\(^2\)=9

b)\(\dfrac{3^{29}.4^{16}}{27^9.8^{11}}\)

=\(\dfrac{3^{29}.2^{32}}{3^{27}.2^{33}}\)

=\(\dfrac{9}{2}\)

Answer 2

\(\dfrac{27^4.4^3}{9^5.8^2}=\dfrac{\left(3^3\right)^4.\left(2^2\right)^3}{\left(3^2\right)^5.\left(2^3\right)^2}=\dfrac{3^{12}.2^6}{3^{10}.2^6}=\dfrac{3^{12}}{3^{10}}=3^2=9\)

_________

\(\dfrac{3^{29}.4^{16}}{27^9.8^{11}}=\dfrac{3^{29}.\left(2^2\right)^{16}}{\left(3^3\right)^9.\left(2^3\right)^{11}}=\dfrac{3^{29}.2^{32}}{3^{27}.2^{33}}=\dfrac{1}{3^2.2}=\dfrac{1}{9.2}=\dfrac{1}{18}\)

Answer 3

Mình sửa phần cuối câu 2:

\(...=\dfrac{3^2}{2}=\dfrac{9}{2}\)