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Ta có :\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^7.\left(3^2\right)^2}{3^3.2^5}\)
\(=\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1}=12\)
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
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\(-\frac{2}{1.3}-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-\frac{2}{9.11}-\frac{2}{11.13}-\frac{2}{13.15}\)
\(=\left(-\frac{2}{1.3}\right)+\left(-\frac{2}{3.5}\right)+\left(-\frac{2}{5.7}\right)+\left(-\frac{2}{7.9}\right)+\left(-\frac{2}{9.11}\right)+\left(-\frac{2}{11.13}\right)+\left(-\frac{2}{13.15}\right)\)
\(=\left(-2\right).\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\right)\)
\(=\left(-2\right).\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\left(-2\right).\left(1-\frac{1}{15}\right)=\left(-2\right).\frac{14}{15}\)
\(=-\frac{28}{15}\)
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BÀI 1
\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}.\)
bài 2
a) \(\frac{1}{2}-\frac{1}{3}+\frac{1}{12}=\frac{6}{12}-\frac{4}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
b) \(\frac{9^9.27^4}{3^8.81^5}=\frac{\left(3^2\right)^9.\left(3^3\right)^4}{3^8.\left(3^4\right)^5}=\frac{3^{18}.3^{12}}{3^8.3^{20}}=\frac{3^{30}}{3^{28}}=3^2=9\)
Study well
Bài 1: \(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}\)
Bài 2:
a)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{12}=\frac{6}{12}-\frac{4}{12}+\frac{1}{12}=\frac{6-4+1}{12}=\frac{1}{4}\)
b)\(\frac{9^9.27^4}{3^8.81^5}=\frac{9^9.3^{12}}{3^8.9^{10}}=\frac{3^4}{9}=\frac{3^4}{3^2}=3^2=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^7.\left(3^2\right)^2}{3^3.2^5}=\frac{2^7.3^4}{3^3.2^5}=2^2.3=12\)
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 \(\frac{2^7.9^2}{3^3.2^5}=12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(\frac{\left(-5\right)^2.20^4}{8^2.\left(-125\right)}=\frac{\left(-5\right)^2.5^4.2^8}{2^6.\left(-5\right)^3}=\left(-5\right)^3.2^2=\left(-125\right).4=-500\)
b, \(\frac{15^{11}.5^7.9^2}{5^{18}.27^6}=\frac{3^{11}.5^{11}.5^7.3^4}{5^{18}.3^{18}}=\frac{3^{15}.5^{18}}{5^{18}.3^{18}}=\frac{1}{3^3}=\frac{1}{27}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Ta có:
\(A=4\frac{25}{16}+25\left(\frac{9}{16}:\frac{125}{64}\right):\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+25.\frac{36}{125}:\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+\frac{36}{5}:\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+\frac{-32}{15}\)
\(\Rightarrow A=\frac{823}{240}\)
Vậy A=.....
b)Ta có:
\(C=\frac{2^3}{3.5}+\frac{2^3}{5.7}+\frac{2^3}{7.9}+...+\frac{2^3}{101.103}\)
\(\Rightarrow C=\frac{2^2.2}{3.5}+\frac{2^2.2}{5.7}+\frac{2^2.2}{7.9}+...+\frac{2^2.2}{101.103}\)
\(\Rightarrow C=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)
\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{101}-\frac{1}{103}\right)\)
\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{103}\right)\)
\(\Rightarrow C=4.\frac{100}{309}\)
\(\Rightarrow C=\frac{400}{309}\)
Vậy C=.....
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{1}{18}\)
\(b,2^8:2^5+3^3.2-12\)
\(=2^3+9.2-12\)
\(=8+18-12\)
\(=26-12\)
\(=14\)
Câu c,d em chưa học nên không biết làm ạ, mong mọi người thông cảm!!!
Sửa lại câu b
\(=2^3+27.2-12\)
\(=8+54-12\)
\(=62-12\)
\(=50\)
\(\frac{2^7.9^2}{3^3.2^5}=\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1.1}=12\)
\(\frac{2^7.9^2}{3^3.2^5}\)
\(=\frac{2^{5+2}.\left(3^2\right)^2}{3^3.2^5}\)
\(=\frac{2^5.2^2.3^4}{3^3.2^5}\)
\(=2^2.3\)
\(=4.3\)
\(=12\)
^_^