a) \(\left(\frac{3}{4}\right)^{45}:\left(\frac{9}{6}\right)^{10}\)

\(=\left(\frac{3}{4}\right)^{45}:\left(\frac{3}{4}\right)^{20}\)

\(=\left(\frac{3}{4}\right)^{25}\)

b) \(\frac{125^{100}.2^{160}}{5^{298}.4^{80}}\)

\(=\frac{5^{300}.2^{160}}{5^{298}.2^{160}}\)

\(=5^2=25\)

a) \(\left(\frac{3}{4}\right)^{45}:\left(\frac{9}{6}\right)^{10}\)

\(\Leftrightarrow\frac{\left(\frac{3}{4}\right)^{45}}{\frac{3^{10}}{2^{10}}}=\frac{\frac{3^{45}}{4^{45}}}{\frac{3^{10}}{2^{10}}}=\frac{3^{45}.2^{10}}{4^{45}}=\frac{3^{35}.2^{10}}{2^{90}}=\frac{3^{35}}{2^{80}}\)

\(\Rightarrow\left(\frac{3}{4}\right)^{45}:\left(\frac{9}{6}\right)^{10}=\frac{3^{35}}{2^{80}}\)

125¹⁰⁰.2¹⁶⁰/5²⁹⁸.4⁸⁰

=5³⁰⁰.2¹⁶⁰/5²⁹⁸.2¹⁶⁰

=5²=25

k mik nha

1. Có 1 thừa số là \(1-\frac{5}{5}=0\) nên tích sẽ bằng 0

2.\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{49}{50}=\frac{1}{50}\)

a, \(\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)

=\(\dfrac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^2+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot\left(3^2\right)^4}\)

=\(\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)

=\(\dfrac{2^{11}\cdot3^6\cdot\left(2^2+3^3\right)}{2^{10}\cdot3^7\cdot\left(2^4+5\cdot3\right)}\)

=\(\dfrac{2^{11}\cdot3^6\cdot31}{2^{10}\cdot3^7\cdot31}\)

=\(\dfrac{2}{3}\)

b, \(\dfrac{\dfrac{8}{27}\cdot\dfrac{9}{16}\cdot\left(-1\right)}{\dfrac{4}{25}\cdot\dfrac{-125}{1728}}\)

=\(\dfrac{\dfrac{8\cdot9\cdot\left(-1\right)}{27\cdot16}}{\dfrac{4\cdot\left(-125\right)}{25\cdot1728}}\)

=\(\dfrac{\dfrac{-1}{6}}{\dfrac{-5}{432}}\)

=\(\dfrac{-1}{6}\cdot\dfrac{-432}{5}\)

=\(\dfrac{72}{5}\)

\(B=\left(\frac{3}{5}\right)^2\cdot5^2-\left(2\frac{1}{4}\right)^3:\left(\frac{3}{4}\right)^3+\frac{1}{2}\)

\(B=\left(\frac{3}{5}\cdot5\right)^2-\left(\frac{9}{4}:\frac{3}{4}\right)^3+\frac{1}{2}\)

\(B=3^2-\left(\frac{9}{4}\cdot\frac{4}{3}\right)^3+\frac{1}{2}\)

\(B=3^2-3^3+\frac{1}{2}=-18+\frac{1}{2}=-\frac{35}{2}\)

2a) \(\frac{3^6+45^4-15^3.4^5}{27^4.25^3+45^6}\)

= \(\frac{3^6+\left(3^2.5\right)^4-\left(3.5\right)^3.\left(2^2\right)^5}{\left(3^3\right)^4.\left(5^2\right)^3+\left(3^2.5\right)^6}\)

= \(\frac{3^6+3^8.5^4-3^3.5^3.4^{10}}{3^{12}.5^6-3^{12}.5^6}=\frac{3^3.\left(3^3+3^5.5^4-5^3.4^{10}\right)}{0}\)(xem lại đề)

b)  \(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{16}{3}\right)^3:\left(\frac{4}{9}\right)^3}{2^7.5^2+512}\)

= \(\frac{\left(\frac{2}{5}.5\right)^7+\left(\frac{16}{3}:\frac{4}{9}\right)^3}{2^7.5^2+2^9}\)

= \(\frac{2^7+12^3}{2^7\left(5^2+2^2\right)}\)

= \(\frac{2^7+\left(2^2.3\right)^3}{2^7.29}\)

= \(\frac{2^7+2^6.3^3}{2^7.29}\)

= \(\frac{2^6\left(1+27\right)}{2^7.29}=\frac{28}{2.29}=\frac{14}{29}\)

mk xin lỗi bn nhé...mk vt nhầm đề...mk vt lại nha:

2a,\(\frac{3^6+45^4-15^3.9^5}{27^4.25^3+45^6}\)