2^3.2^2+7.9
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\(\frac{2^7.9^2}{3^3.2^5}\)\(=\frac{2^7.\left(3^2\right)^2}{3^3.2^5}=\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1}=2^2.3=4.3=12\)
\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^7.\left(3^2\right)^2}{3^3.2^5}=\frac{2^7.3^4}{3^3.2^5}=2^2.3=12\)
Có gì sai sót thì bỏ qua nhé chị !
Ta có : \(\frac{2^7.9^2}{3^3.2^5}=\frac{2^5.2^2.81}{27.2^5}=\frac{2^5.2^2.27.3}{27.2^5}=2^2.3=12\)
Vậy \(\frac{2^7.9^2}{3^3.2^5}=12\)
\(\dfrac{2^7.9^2}{3^3.2^5}\)
\(=\dfrac{2^7.\left(3^2\right)^2}{3^3.2^5}\)
\(=\dfrac{2^7.3^4}{3^3.2^5}\)
\(=2^2.3\)
\(=4.3\)
\(=12\)
\(\dfrac{2^7.9^2}{3^3.2^5}=\dfrac{2^7.\left(3^2\right)^2}{3^3.2^5}=\dfrac{2^7.3^4}{3^3.2^5}=\dfrac{2^2.3}{1}=4.3=12\)
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
\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1.1}=12\)
\(a)\) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{197.198}+\frac{1}{198.199}\right)\)
\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)
\(A=\frac{1}{199}-\left(1-\frac{1}{199}\right)\)
\(A=\frac{1}{199}-1+\frac{1}{199}\)
\(A=\frac{-197}{199}\)
Chúc bạn học tốt ~
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 ! >_<
=32+63=95