\(\frac{\left(-2\right)^3\cdot3^3\cdot5^3\cdot7\cdot8}{3\cdot2^4\cdot5^3\cdot14}\)

\(=\frac{2^2\cdot\left(-2\right)\cdot3^3\cdot5^3\cdot7\cdot2^3}{3\cdot2^4\cdot5^3\cdot7\cdot2}=\frac{\left(-2\right)\cdot3^2}{1}=-18\)

\(\frac{\left(2^{17}+5^{17}\right)\left(3^{14}-5^{12}\right)\left(2^4-4^2\right)}{15^2+5^3+67^7}=\frac{\left(2^{17}+5^{17}\right)\left(3^{14}-5^{12}\right).0}{15^2+5^3+67^7}=0\)

giúp mk vs

1. \(\frac{9}{30}=\frac{3}{10};\frac{98}{80}=\frac{49}{40};\frac{15}{1000}=\frac{3}{200}\)

Vì \(200⋮10;200⋮40\) 

=> BCNN(10; 40; 200) = 200

200 : 10 = 20

200 : 40 = 5

=> \(\frac{3}{10}=\frac{3\cdot20}{10\cdot20}=\frac{60}{200}\), \(\frac{49}{40}=\frac{49\cdot5}{40\cdot5}=\frac{245}{200}\)

Câu a:

\(\frac{2^{12}.3^5-4^6.9^2}{12^6+8^4.3^5}\)

= \(\frac{2^{12}.3^5-2^{12.3^6}}{2^{12}.3^6+2^{12}.3^5}\)

= \(\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}\)

= \(\frac{2}{3.4}\)

= \(\frac16\)

Câu b:

\(\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

= \(\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)

= \(\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+8\right)}\)

= \(\frac{5.\left(-6\right)}{9}\)

=- 10/3