\(\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)\(\dfrac{ }{ }\)
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1 tháng 5 2015
\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}=\frac{2.2.2.5.7.5.5.7.7.7}{2.5.7.7.2.5.7.7}=\frac{2.5}{1}=10\)
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1 tháng 5 2015
\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}=\frac{2^3.\left(5.5^2\right).\left(7.7^3\right)}{2^2.5^2.7^{2^2}}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=2.5=10\)
17 tháng 5 2016
E = \(\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=\frac{2^2.2.5^2.5.7^4}{2^2.5^2.7^4}=2.5=10\)
17 tháng 5 2016
\(E=\frac{2^3.5.7.5^2.7^3}{2^2.5^2.7^4}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=2.5=10\)
2
5 tháng 5 2022
`a)1/2 . [-3]/4 . [-5]/8 . [-8]/9=[1. (-3).(-5).(-8)]/[2.4.8.3.3]=[-5]/[2.4.3]=[-5]/24`
`b)(2/[1.3]+2/[3.5]+2/[5.7]).([10.13]/3-[2^2]/3-[5^3]/3)`
`=(1-1/3+1/3-1/5+1/5-1/7).[10.13-2^2-5^3]/3`
`=(1-1/7).[130-4-125]/3`
`=6/7 . 1/3 = 2/7`
____________________________________________________
`8/9+1/9 . 2/9+1/9 . 7/9`
`=8/9+1/9.(2/9+7/9)`
`=8/9+1/9 . 9/9`
`=8/9+1/9=9/9=1`
5 tháng 5 2022
a) \(\dfrac{1}{2}\cdot\dfrac{-3}{4}\cdot\dfrac{-5}{8}\cdot\dfrac{-8}{9}\)
\(=\dfrac{1\cdot\left(-3\right)\cdot\left(-5\right)\cdot\left(-8\right)}{2\cdot4\cdot8\cdot9}\)
\(=-\dfrac{5}{24}\)
b) \(\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}\right)\cdot\left(\dfrac{10\cdot13}{3}-\dfrac{2^2}{3}-\dfrac{5^3}{3}\right)\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}\right)\cdot\left(\dfrac{130}{3}-\dfrac{4}{3}-\dfrac{125}{3}\right)\)
\(=\left(1-\dfrac{1}{7}\right)\cdot\dfrac{1}{3}\)
\(=\dfrac{6}{7}\cdot\dfrac{1}{3}\)
\(=\dfrac{2}{7}\)
\(\dfrac{8}{9}+\dfrac{1}{9}\cdot\dfrac{2}{9}+\dfrac{1}{9}\cdot\dfrac{7}{9}\)
\(=\dfrac{8}{9}+\dfrac{2}{81}+\dfrac{7}{81}\)
\(=\dfrac{72}{81}+\dfrac{2}{81}+\dfrac{7}{81}\)
\(=1\)
24 tháng 5 2022
Bài 2:
a: \(=44\cdot82-400+18\cdot44\)
\(=44\cdot100-400=4400-400=4000\)
b: \(=6^2:\left\{780:\left[390-125\cdot49+65\right]\right\}\)
\(=36:\left\{780:\left[-5670\right]\right\}\)
\(=36:\dfrac{-26}{189}=\dfrac{-3402}{13}\)
27 tháng 8 2019
a)
\(=\frac{3}{5}.\frac{3}{7}+\frac{3}{5}.\frac{4}{7}-\left(1+\frac{3}{5}\right)\)
\(=\frac{3}{5}\left(\frac{3}{7}+\frac{4}{7}\right)-1-\frac{3}{5}\)
\(=\frac{3}{5}-1-\frac{3}{5}\)
\(=-1\)
b) \(=\frac{2^2.5.7.5^2.7^3}{2^2.5^2.7^{2.2}}\)
\(=\frac{2^2.5^{1+2}.7^{3+1}}{2^2.5^2.7^4}=\frac{2^2.5^3.7^4}{2^2.5^2.7^4}=2^{2-2}.5^{3-2}.7^{4-4}=2^0.5^1.7^0=1.5.1=5\)
Đặt A=\(\dfrac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
A=\(\dfrac{2^3.5.7.5^2.7^3}{2^2.5^2.7^4}\)
A=\(\dfrac{2^3.5^3.7^4}{2^2.5^2.7^4}\)
A=2.\(5^2\)
A=2.25
A=50