Ta có :\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^7.\left(3^2\right)^2}{3^3.2^5}\)

\(=\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1}=12\)

Trả lời:

\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^2.\left(3^2\right)^2}{3^3}=\frac{2^2.3^4}{3^3}=\frac{2^2.3}{1}=4.3=12\)

Học tốt

\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1.1}=12\)

\(\frac{2^7.9^2}{3^3.2^5}\)

\(=\frac{2^{5+2}.\left(3^2\right)^2}{3^3.2^5}\)

\(=\frac{2^5.2^2.3^4}{3^3.2^5}\)

\(=2^2.3\)

\(=4.3\)

\(=12\)

^_^

\(\frac{2^7\cdot9^2}{3^3\cdot2^5}\)

⇒\(\frac{2^7\cdot\left(3^2\right)^2}{3^3\cdot2^5}\)

⇒\(\frac{2^7\cdot3^4}{3^3\cdot2^5}\)

⇒\(2^2\cdot2\)

⇒\(4\cdot2\)

➩\(8\)

Chúc bạn học tốt nhé Thuong Hoai!

\(\frac{2^7.9^2}{3^3.2^5}\)\(=\frac{2^7.\left(3^2\right)^2}{3^3.2^5}=\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1}=2^2.3=4.3=12\)

\(\dfrac{2^7.9^2}{3^3.2^5}\)

\(=\dfrac{2^7.\left(3^2\right)^2}{3^3.2^5}\)

\(=\dfrac{2^7.3^4}{3^3.2^5}\)

\(=2^2.3\)

\(=4.3\)

\(=12\)

\(\dfrac{2^7.9^2}{3^3.2^5}=\dfrac{2^7.\left(3^2\right)^2}{3^3.2^5}=\dfrac{2^7.3^4}{3^3.2^5}=\dfrac{2^2.3}{1}=4.3=12\)

\(\dfrac{2^7.9^2}{3^3.2^5}=\dfrac{2^7.\left(3^2\right)^2}{3^3.2^5}=\dfrac{2^2.3^4}{3^3}=2^2.3=4.3=12\)

=32+63=95

\(a)\) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{197.198}+\frac{1}{198.199}\right)\)

\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-\left(1-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-1+\frac{1}{199}\)

\(A=\frac{-197}{199}\)

Chúc bạn học tốt ~ 

làm hộ mk lun câu b ik

a) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...\frac{1}{197.198}+\frac{1}{198.199}\right)\)

\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-(1-\frac{1}{199})\)

\(A=\frac{1}{199}-1+\frac{1}{199}\)

\(A=\frac{-197}{199}\)

Vậy \(A=\frac{-197}{199}\)

b) cũng thế 

Mk sửa đề nha :

\(C=\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+\frac{12}{7}\)

   \(=\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)

   \(=\frac{-5}{7}.1+\frac{12}{7}\)

   \(=\frac{-5}{7}+\frac{12}{7}\)

   \(=1\)

Study well ! >_<