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\(\frac{4^{1007}.9^{1007}}{3^{2015}.16^{503}}=\frac{4^{1007}.\left(3^2\right)^{1007}}{3^{2015}.\left(4^2\right)^{503}}=\frac{4^{1007}.3^{2014}}{3^{2015}.4^{1006}}=\frac{4}{3}\)
\(\frac{4^{1007}.9^{1007}}{3^{2015}.2^{2016}}=\frac{\left(2^2\right)^{1007}.\left(3^2\right)^{1007}}{3^{2015}.2^{2016}}\)
\(=\frac{2^{2014}.3^{2014}}{3^{2015}.2^{2016}}=\frac{2^{2014}.3^{2014}}{3^{2014}.2^{2014}.3.2^2}\)
\(=\frac{1}{3.2^2}=\frac{1}{3.4}=\frac{1}{12}\)
Rút gọn
\(\frac{4^{1007}\cdot9^{1007}}{3^{2015}\cdot2^{2016}}=\frac{\left(2^2\right)^{2007}\cdot\left(3^2\right)^{1007}}{3^{2015}\cdot2^{2016}}\)
\(=\frac{2^{2\cdot1007}\cdot3^{2\cdot1007}}{3^{2015}\cdot2^{2016}}=\frac{2^{2014}\cdot3^{2014}}{3^{2015}\cdot2^{2016}}\)
\(=\frac{1}{3.2^2}=\frac{1}{12}\)
Vậy ...
hok tót .
Câu 1 :
\(\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^5}\)
= \(\frac{5^{32}.2^{86}.5^{43}}{\left(-2\right)^{87}.5^{15}}\)
= \(\frac{5^{72}.\left(-2\right)^{86}}{\left(-2\right)^{87}.5^{75}}\)
= \(\frac{1}{-2}\)
Câu 2 :
\(\frac{5^4.18^4}{125.9^5.16}\)
= \(\frac{5^4.2^4.3^8}{5^3.3^{10}.2^4}\)
= \(\frac{5}{3^2}\)
= \(\frac{5}{9}\)
Câu 3 :
\(\frac{9^{18}.2^{29}}{8^9.27^{12}}\)
= \(\frac{3^{36}.2^{29}}{2^{27}.3^{36}}\)
= \(2^2\)
= 4
Bài giải
\(\frac{2-x}{2015}+\frac{3-x}{1007}+\frac{4-x}{671}=\frac{2005-x}{2}\)
\(( \frac{2-x}{2015}+1 )+ (\frac{3-x}{1007}+2 )+ ( \frac{4-x}{671}+3 )=\frac{2005-x}{2}+6\)
\(\frac{2017-x}{2015}+\frac{2017-x}{1007}+\frac{2017-x}{671}=\frac{2017-x}{2}\)
\(\frac{2017-x}{2015}+\frac{2017-x}{1007}+\frac{2017-x}{671}-\frac{2017-x}{2}=0\)
\((2017-x)(\frac{1}{2015}+\frac{1}{1007}+\frac{1}{671}-\frac{1}{2})=0\)
Do \(\frac{1}{2015}+\frac{1}{1007}+\frac{1}{671}-\frac{1}{2}\ne0\)
\(\Rightarrow\text{ }2017-x=0\)
\(\Rightarrow\text{ }x=2017\)
Cho hỏi viết số mũ như thế nào vậy Chỉ cho mk với
\(\frac{4^{1007}\cdot9^{1007}}{3^{2015}\cdot16^{503}}\)
\(=\frac{\left(2^2\right)^{1007}\cdot\left(3^2\right)^{1007}}{3^{2015}\cdot\left(2^4\right)^{503}}\)
\(=\frac{2^{2014}\cdot3^{2014}}{3^{2015}\cdot2^{2012}}\)
\(=\frac{2^2\cdot1}{3\cdot1}=\frac{4}{3}\)