bài 1 tính : \(\frac{2^5.7+2^5.59}{2^5.5^2-2^5.3}\)
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29 tháng 3 2019
\(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}=\frac{8}{22}\)
29 tháng 3 2019
\(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}\)
\(\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}\)=\(\frac{8}{22}=\frac{4}{11}\)
Hok tốt
8 tháng 2 2017
Ta có: \(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}\) = \(\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}\) = \(\frac{2^5.8}{2^5.22}\) = \(\frac{8}{22}\) =\(\frac{56}{154}\)
\(\frac{3^4.5-3^6}{3^4.13+3^4}\) = \(\frac{3^4.\left(5-3^2\right)}{3^4.\left(13+1\right)}\) = \(\frac{3^4.\left(-4\right)}{3^4.14}\) = \(\frac{-4}{14}\)= \(\frac{-44}{154}\)
12 tháng 4 2018
Câu a, b phân k ra là ok
\(c)\) \(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(A=\left(-1\right)\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(A=\left(-1\right)\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=\left(-1\right)\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\left(-1\right)\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(A=\left(-1\right).\frac{3}{20}\)
\(A=\frac{-3}{20}\)
Vậy \(A=\frac{-3}{20}\)
Chúc bạn học tốt ~
11 tháng 3 2017
\(a,\left(-25\right).\left(75-45\right)-75.\left(45-25\right)\)
\(=\left(-25\right).20-75.20\)
\(=20.\left(-25-75\right)\)
\(=20.\left(-100\right)\)
\(=-2000\)
\(b,\frac{18.12-48.15}{-3.270-3.30}\)
\(=\frac{18.12-12.4.15}{-3.270-3.30}\)
\(=\frac{12.\left(18-60\right)}{-3.\left(270+30\right)}\)
\(=\frac{12.\left(-42\right)}{-3.300}\)
\(=\frac{14}{25}\)
\(c,\frac{2^5.7+2^5}{2^5.5^2-2^5.3}\)
\(=\frac{2^5.\left(7+1\right)}{2^5\left(25-3\right)}\)
\(=\frac{2^5.8}{2^5.22}\)
\(=\frac{8}{22}\)
\(=\frac{4}{11}\)
TV
23 tháng 3 2018
a , 5.6 + 5.7 / 5.8 + 20 = 5.6 + 5.7 / 5.8 + 5 . 4 = 5 . ( 6+7 ) / 5 . ( 8 + 4 ) = 6 + 7 / 8 + 4 = 13 / 12 8 . 9 + 4 .15 / 12 . 7 - 180 = 4 . 2 . 3 . 3 + 2 . 2 . 3 . / 4 . 3 . 7 - 180 = 4 . 2 . 3 . 3 + 2.2.3.5 / 3 . 4 . 7 - 3 . 2 . 2 . 3. 5 = 1 . 2 . 1 . 1 + 1 . 1 . 1. 1 / 1 . 1 . 7 - 1 . 1 . 1 . 1 .1 = 3 / 6 = 1/2 b , 2^5 . 7 +2^5 / 2^5 . 5^2 - 2^5 .3 = 2^5 . ( 7 + 1) / 2^5 ( 5^2 - 3 ) = 7+1 / 5^2 - 3 = 8 / 22 = 4 / 11 3^4 . 5 - 3^6 / 3^4 . 13 + 3^4 = 3^4 . 5 - 3^4 . 3^2 / 3^4 . 13 + 3^4 = 3^4 . ( 5 - 3^2 ) / 3^4 . ( 13 + 1 ) = 5 - 3^2 / 13 + 1 = -4 / 14 = -2 / 12
TB
1
29 tháng 7 2017
\(\frac{2^5.7+2^5}{2^5+5^2-2^5.3}\)
\(=\frac{2^5.8}{5^2.3}\)
\(=\frac{32.8}{25.3}\)
\(=\frac{256}{75}\)
22 tháng 2 2015
a) \(\frac{25.9-25.17}{-8.80-8.10}=\frac{25.\left(9-17\right)}{-8.\left(80+10\right)}=\frac{25.\left(-8\right)}{-8.90}=\frac{5}{18}\)
b) \(\frac{48.12-48.15}{-3.270-3.30}=\frac{48.\left(12-15\right)}{-3.\left(270+30\right)}=\frac{48.\left(-3\right)}{-3.300}=\frac{4}{25}\)
c) \(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}=\frac{2^5.8}{2^5.\left(25-3\right)}=\frac{2^5.8}{2^5.22}=\frac{4}{11}\)
d) \(\frac{3^4.5-3^6}{3^4.13+3^4}=\frac{3^4.\left(5-3^2\right)}{3^4.\left(13+1\right)}=\frac{3^4.\left(5-9\right)}{3^4.14}=\frac{3^4.\left(-4\right)}{3^4.14}=\frac{-2}{7}\)
\(\frac{2^5.7+2^5.59}{2^5.5^2-2^5.3}\)
=\(\frac{\left(2^5\right).\left(7+59\right)}{\left(2^5\right).\left(5^2-3\right)}\)
=\(\frac{7+59}{5^2-3}\)
=\(\frac{64}{22}\)
=\(\frac{32}{11}\)