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Ta có:
\(\frac{2^2.2.3^2.3^2}{2^2.3^2.5}\) =\(\frac{2.3^2}{5}\) =\(\frac{18}{5}\)
Vậy
Ta cã: \(\frac{2^3.3^4}{2^2.3^2.5}\)
\(=\frac{2^2.2.3^2.3^2}{2^2.3^2.5}\)
\(=\frac{2.3^2}{5}\)
\(=\frac{18}{5}\)
\(\frac{2^3.3^4}{2^2.3^2.5}=\frac{2^2.3^2.2.3^2}{2^2.3^2.5}=\frac{2^2.3^2.18}{2^2.3^2.5}=\frac{18}{5}\)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
= \(1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-...-\left(\frac{1}{98}-\frac{1}{98}\right)-\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)
= \(1-\frac{1}{100}\)
= \(\frac{99}{100}\)
Vậy ...
B = \(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)
= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
= \(\frac{1}{2}-\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{1}{8}-\frac{1}{8}\right)-...-\left(\frac{1}{17}-\frac{1}{17}\right)-\frac{1}{20}\)
= \(\frac{1}{2}-\frac{1}{20}\)
= \(\frac{9}{20}\)
Vậy B = 9/20
\(\frac{2^4\cdot5^2\cdot11^2\cdot7}{2^3\cdot5^3\cdot7^2\cdot11}\)
\(=\frac{2\cdot11}{5\cdot7}\)
\(=\frac{22}{35}\)
TA CÓ : \(\frac{2^3.3^4}{2^2.3^2.5}\)= \(\frac{2^3.3^4}{\left(2.3\right)^2.5}\)= \(\frac{2^3.3^4}{6^2.5}\)= \(\frac{2^3.3^4}{36.5}\)= \(\frac{8.81}{180}\)= \(\frac{648}{180}\)= 648 : 180 = 3,6 HOẶC \(\frac{648}{180}\)= \(\frac{18}{5}\)
\(\frac{2^3\cdot3^4}{2^2\cdot3^2\cdot5}=\frac{2^2\cdot2\cdot3^4}{2^2\cdot3^2\cdot5}=\frac{2\cdot3^4}{3^2\cdot5}=\frac{2\cdot3^2}{5}=\frac{18}{5}\)
\(\frac{2^3.3}{2^2.3^2.5}=\frac{2^2.2^1.3}{2^2.3.3.5}=\frac{2}{3.5}=\frac{2}{15}\)
a)\(\frac{121212}{131313}=\frac{121212:10101}{131313:10101}=\frac{12}{13}\)
b)\(\frac{2^3\cdot3^4}{2^2\cdot3^2\cdot5}=\frac{2^2\cdot3^2\cdot3^2.2}{2^2\cdot3^2\cdot5}=\frac{3^2\cdot2}{5}=\frac{9\cdot2}{5}=\frac{18}{5}\)
=2.3^2/5
=2.9/5
=18/8
Bài làm
2³.3⁴/2² . 3² . 5
= 2 . 3²/5
= 2 . 9 / 5
= 18 / 5
Vậy giá trị của biểu thức trên là 18/5