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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 \(\frac{648}{180}\)= \(\frac{18}{5}\)
Ta có:
\(\frac{2^3.3^4}{2^2.3^2.5}=\frac{2.3^2}{5}=\frac{18}{5}\)
\(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}=\frac{2.11}{5.7}=\frac{22}{35}\)
a: \(=\dfrac{-4\cdot13\cdot9\cdot5}{3\cdot4\cdot5\cdot2\cdot13}=\dfrac{3}{2}\)
b: \(=\dfrac{1}{2}\cdot\dfrac{1}{3}\cdot5=\dfrac{5}{6}\)
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A.\(\frac{2^3.9^2}{2^2.3^2.5}=\frac{2^2.2.\left(3^2\right)^2}{2^2.3^2.5}=\frac{2.3^2}{5}=\frac{18}{5}\)
B.\(\frac{2^4.5^5.11^2.7}{2^3.5^3.7^2.11}=\frac{2^3.2.5^3.5^2.11.11.7}{2^3.5^3.7.7.11}=\frac{2.5^2.11}{7}=\frac{61}{7}\)
a)
\(y=\frac{2^3.3^4}{2^2.3^2.5}=\frac{2^2.2.3^2.3^2}{2^2.3^2.5}=\frac{2.3^2}{5}=\frac{18}{5}\)
b)\(y=\frac{1989.1990+3978}{1992.1991-3984}=\frac{1989.1990+1989.2}{1992.1991-1992.2}=\frac{1989.1992}{1992.1989}=1\)(vì tử bằng mẫu)
\(a.\frac{2\cdot\left(-13\right)\cdot9\cdot10}{\left(-3\right)\cdot4\cdot\left(-5\right)\cdot26}\)
\(=\frac{2\cdot\left(-13\right)\cdot3\cdot3\cdot2\cdot5}{\left(-3\right)\cdot2\cdot2\cdot\left(-5\right)\cdot13\cdot2}\)
\(=-\frac{3}{2}\)
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) \(\frac{1998\cdot1990+3978}{1992\cdot1991-3984}\)