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25 tháng 3 2020
\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^7.\left(3^2\right)^2}{3^3.2^5}=\frac{2^7.3^4}{3^3.2^5}=2^2.3=12\)
25 tháng 3 2020
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 \(\frac{2^7.9^2}{3^3.2^5}=12\)
NH
2
HN
19 tháng 12 2018
Ta có :\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^7.\left(3^2\right)^2}{3^3.2^5}\)
\(=\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1}=12\)
19 tháng 12 2018
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
DH
27 tháng 12 2019
\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1.1}=12\)
28 tháng 11 2017
- bạn khác chọn (k) đúng cho mình
- Chỉ ghi đáp số mà không có lời giải, hoặc nội dung không liên quan đến câu hỏi.
1 tháng 8 2018
Bài 1
\(\frac{2^7.9^2}{3^3.2^5}\)
\(=\frac{2^7.3^4}{3^3.2^5}\)
\(=2^2.3\)
\(=12\)
25 tháng 2 2020
\(\frac{2^7.9^2}{3^3.2^5}\)\(=\frac{2^7.\left(3^2\right)^2}{3^3.2^5}=\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1}=2^2.3=4.3=12\)
MV
3 tháng 8 2017
a,
\(\dfrac{4^2\cdot4^3}{2^{10}}=\dfrac{4^5}{2^{10}}=\dfrac{\left(2^2\right)^5}{2^{10}}=\dfrac{2^{10}}{2^{10}}=1\)
b,
\(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,2\cdot3\right)^5}{\left(0,2\right)^5\cdot0,2}=\dfrac{\left(0,2\right)^5\cdot3^5}{\left(0,2\right)^5\cdot0,2}=\dfrac{243}{0,2}=\dfrac{243}{\dfrac{1}{5}}=243\cdot5=1215\)
c,
\(\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}=\dfrac{2^6\cdot2\cdot3^6}{2^5\cdot3^5\cdot2^6}=\dfrac{3}{2^4}=\dfrac{3}{16}\)
d,
\(\dfrac{6^3+3\cdot6^2+3^3}{-13}=\dfrac{\left(2\cdot3\right)^3+3\cdot\left(2\cdot3\right)^2+3^3}{-13}=\dfrac{2^3\cdot3^3+3\cdot2^2\cdot3^2+3^3}{-13}=\dfrac{2^3\cdot3^3+2^2\cdot3^3+3^3}{-13}\dfrac{3^3\left(2^3+2^2+1\right)}{-13}=\dfrac{3^3\cdot13}{-13}=-3^3=-27\)
SA
17 tháng 11 2018
\(a,2^3.2^2+3^4.3-4^2\)
\(=8.4+81.3-16\)
\(=32+243-16\)
\(=259\)
\(b,\left(\dfrac{-2}{3}+\dfrac{3}{4}\right)^2:\dfrac{5}{11}\)
\(=\left(\dfrac{-8}{12}+\dfrac{9}{12}\right)^2.\dfrac{11}{5}\)
\(=\left(\dfrac{1}{12}\right)^2.\dfrac{11}{5}\)
\(=\dfrac{1}{144}.\dfrac{11}{5}\)
\(=\dfrac{11}{720}\)
NA
1
TT
22 tháng 9 2017
\(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|-3.2+\dfrac{2}{5}\right|\)
\(\Rightarrow\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|\dfrac{-28}{5}\right|\)
\(\Rightarrow\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\dfrac{28}{5}\)
\(\Rightarrow\left|x-\dfrac{1}{3}\right|=\dfrac{28}{5}-\dfrac{4}{5}=\dfrac{24}{5}\)
\(\Rightarrow x-\dfrac{1}{3}=\left(\pm\dfrac{24}{5}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{24}{5}\\x-\dfrac{1}{3}=\dfrac{-24}{5}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{24}{5}+\dfrac{1}{3}\\x=\dfrac{-24}{5}+\dfrac{1}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{77}{15}\\x=\dfrac{-67}{15}\end{matrix}\right.\)
Vậy...
QN
16 tháng 9 2017
a) \(A=\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{72}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=\left(\dfrac{5}{15}+\dfrac{9}{15}+\dfrac{1}{15}\right)-\left(\dfrac{27}{36}+\dfrac{8}{36}+\dfrac{1}{36}\right)+\dfrac{1}{72}\)
\(=1-1+\dfrac{1}{72}\)
\(=0+\dfrac{1}{72}=\dfrac{1}{72}\)
b) \(B=\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{2}{9}+\dfrac{7}{13}-\dfrac{2}{11}-\dfrac{5}{9}+\dfrac{3}{7}-\dfrac{1}{5}\)
\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{5}{9}-\dfrac{5}{9}\right)-\left(\dfrac{2}{9}-\dfrac{7}{13}+\dfrac{2}{11}\right)\)
\(=0+0+0-\left(\dfrac{286}{1287}-\dfrac{693}{1287}+\dfrac{234}{1287}\right)\)
\(=-\left(-\dfrac{173}{1287}\right)\)
\(=\dfrac{173}{1287}\)
c) \(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-.....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{-49}{50}\)
\(\dfrac{2^7.9^2}{3^3.2^5}\)
\(=\dfrac{2^7.\left(3^2\right)^2}{3^3.2^5}\)
\(=\dfrac{2^7.3^4}{3^3.2^5}\)
\(=2^2.3\)
\(=4.3\)
\(=12\)
\(\dfrac{2^7.9^2}{3^3.2^5}=\dfrac{2^7.\left(3^2\right)^2}{3^3.2^5}=\dfrac{2^7.3^4}{3^3.2^5}=\dfrac{2^2.3}{1}=4.3=12\)