`@` `\text {Ans}`

`\downarrow`

\(\dfrac{21^2\cdot14\cdot125}{35^2\cdot125}\)

`=`\(\dfrac{3^2\cdot7^2\cdot2\cdot7\cdot5^2}{5^2\cdot7^2\cdot5^2}\)

`=`\(\dfrac{3^2\cdot2\cdot7\cdot5^2\cdot7^2}{5^2\cdot5^2\cdot7^2}\)

`=`\(\dfrac{3^2\cdot2\cdot7}{5^2}=\dfrac{126}{25}\)

B=2^18/2^12=2^18:2^12=2^6

\(A=\frac{4^6.3^4.9^5}{6^{12}}=\frac{\left(2^2\right)^6.3^3.\left(3^2\right)^5}{6^{12}}\)

\(=\frac{2^{12}.3^3.3^{10}}{6^{12}}=3^{13}.3^{12}=3^{25}\)

\(A=\frac{4^6.3^4.9^5}{6^{12}}\)

\(A=\frac{2^6.2^6.3^4.3^5.3^5}{2^{12}.3^{12}}\)

\(A=\frac{3^3.3^5}{1}\)

\(A=3^8\)
 

\(B=\frac{21^2.14.125}{35^3.6}\)

\(B=\frac{3^2.7^2.2.7.5^3}{5^3.7^3.2.3}\)

\(B=\frac{3.1.1.1.1}{1.1.1.1}\)

\(B=3\)

\(\dfrac{1.3.5+2.6.10+4.12.20+7.21.35}{1.5.7+2.10.14+4.20.28+7.35.49}\)

\(=\dfrac{1.3.5+2^3.1.3.5+2^6.1.3.5+7^3.1.3.5}{1.5.7+2^3.1.5.7+2^6.1.5.7+7^3.1.5.7}\)

\(=\dfrac{1.3.5\left(1+2^3+2^6+7^3\right)}{1.5.7\left(1+2^3+2^6+7^3\right)}\)

\(=\dfrac{1.3.5}{1.5.7}\)

giúp mk vs các bn

\(\frac{2.6.10+6.10.14+10.14.18+...+194.198.202}{1.3.5+3.5.7+...+97.99.101}\)

\(=\frac{2^3.1.3.5+2^3.3.5.7+2^3.97.99.101}{1.3.5+3.5.7+...+97.99.101}\)

\(=\frac{2^3\left(1.3.5+3.5.7+...+97.99.101\right)}{1.3.5+3.5.7+...+97.99.101}\)

\(=\frac{2^3}{1}=8\)

Vậy A = 8

a.\(\frac{3\cdot4\cdot7}{12\cdot8\cdot9}\)= \(\frac{3\cdot4\cdot7}{3\cdot4\cdot8\cdot9}\)= \(\frac{7}{72}\) 

b. \(\frac{4\cdot5\cdot6}{12\cdot10\cdot8}\)= \(\frac{4\cdot5\cdot2\cdot3}{3\cdot4\cdot5\cdot2\cdot8}\)= \(\frac{1}{8}\) 

c.\(\frac{5\cdot6\cdot7}{12\cdot14\cdot15}\)= \(\frac{5\cdot6\cdot7}{2\cdot6\cdot2\cdot7\cdot3\cdot5}\)= \(\frac{1}{12}\)

\(\frac{\cdot}{ }\)là dấu * ;là dấu nhân

a, \(\frac{3.4.7}{12.8.9}\)= \(\frac{3.4.7}{3.4.8.9}\)= \(\frac{7}{72}\)

b, \(\frac{4.5.6}{12.10.8}\)= \(\frac{4.5.6}{3.4.2.5.8}\)= \(\frac{1}{8}\)

c, \(\frac{5.6.7}{12.14.15}\)= \(\frac{5.6.7}{2.6.2.7.3.5}\)= \(\frac{1}{12}\)