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\(\frac{2^35^27^23^7}{49.5^3.3^6.11}\)
= \(\frac{2^31^21^21^7}{7.1^3.1^6.11}\)
= \(\frac{8}{77}\)
\(\frac{2^3.5^2.7^2.3^7}{49.5^3.3^6.11}\)\(=\frac{2^3.5^2.7^2.3^7}{7^2.5^3.3^6.11}=\frac{2^3.3}{5.11}=\frac{24}{55}\)
Ủng hộ mk nha!!!
Chúc các bn học tốt nha !!
a. \(\dfrac{2^3.5^2.7^2.3^7}{49.5^3.3^6.11}\)
= \(\dfrac{2^3.3}{5.11}=\dfrac{24}{55}\)
b. \(4.\left(\dfrac{-1}{2}\right)^3-2.\left(\dfrac{-1}{2}\right)^2+3\left(\dfrac{-1}{2}\right)+1\)
=\(-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}\)
= 3\(\left(\dfrac{-1}{2}\right)\)
=\(\dfrac{-3}{2}\)
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 \(\frac{648}{180}\)= \(\frac{18}{5}\)
\(a.\frac{2\cdot\left(-13\right)\cdot9\cdot10}{\left(-3\right)\cdot4\cdot\left(-5\right)\cdot26}\)
\(=\frac{2\cdot\left(-13\right)\cdot3\cdot3\cdot2\cdot5}{\left(-3\right)\cdot2\cdot2\cdot\left(-5\right)\cdot13\cdot2}\)
\(=-\frac{3}{2}\)
b) \(\frac{2^3\cdot3^4}{2^2\cdot3^2\cdot5}=\frac{2\cdot3^2}{5}=\frac{2\cdot9}{5}=\frac{18}{5}\)
\(\frac{2^4\cdot5^2\cdot11^2\cdot7}{2^3\cdot5^3\cdot7^2\cdot11}=\frac{2\cdot1\cdot11\cdot1}{1\cdot5\cdot7\cdot1}=\frac{22}{35}\)
c) \(\frac{121\cdot75\cdot130\cdot169}{39\cdot60\cdot11\cdot198}=\frac{11\cdot11\cdot13\cdot10\cdot169}{13\cdot3\cdot6\cdot10\cdot11\cdot11\cdot6\cdot3}\)
\(=\frac{169}{3\cdot6\cdot6\cdot3}=\frac{169}{324}\)
d) \(\frac{1998\cdot1990+3978}{1992\cdot1991-3984}\)
a) \(\frac{3024-12}{5292-21}=\frac{3012}{5271}=\frac{4}{7}\)
b) \(\frac{2^3.3}{2^2.3^2.5}=\frac{2^1.1}{1.3^1.5}=\frac{2}{15}\)
c) dễ mà đợi mk tính rùi làm sau
duyệt đi
c) = \(\frac{2^1.5^1.1}{1.1.7^1.11}=\frac{10}{77}\)
ko bít đúng ko
duyệt đi
\(\frac{2^2.3^3.5^7}{2^3.3^4.5^6}=\frac{1.1.5}{2.3.1}=\frac{5}{6}\)
Làm tương tự
\(B=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....++\frac{1}{9}-\frac{1}{10}\)
\(B=1-\frac{1}{10}=\frac{9}{10}\)
\(C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(C=1-\frac{1}{100}\)
\(C=\frac{99}{100}\)
a) 2,25 : \(\frac{-6}{25}=\frac{9}{4}\cdot\frac{-25}{6}=\frac{9\cdot\left(-25\right)}{4\cdot6}=\frac{3\cdot3\cdot\left(-25\right)}{4\cdot3\cdot2}=\frac{-75}{8}\)
b) \(\frac{2^3\cdot5^2\cdot7^2\cdot3^7}{49\cdot5^3\cdot3^6\cdot11}=\frac{2^3\cdot5^2\cdot7^2\cdot3^6\cdot3}{7^2\cdot5^2\cdot5\cdot3^6\cdot11}=\frac{2^3\cdot3}{5\cdot11}=\frac{24}{55}\)