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Khách
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10 tháng 9 2015
C1: \(A=\left(\frac{36}{6}-\frac{4}{6}+\frac{3}{6}\right)-\left(\frac{150}{30}+\frac{50}{30}-\frac{45}{30}\right)-\left(\frac{18}{6}-\frac{14}{6}+\frac{15}{6}\right)\)
\(=\frac{35}{6}-\frac{155}{30}-\frac{19}{6}=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=-\frac{15}{6}=-2\frac{1}{2}\)
C2: \(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(=\left(6-5-3\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
\(=-2-0-\frac{1}{2}=-2\frac{1}{2}\)
CM
25 tháng 9 2017
A=\(\dfrac{12^3\cdot8^3\cdot5}{2^9\cdot\left(3\cdot5\right)^2}\)
A=\(\dfrac{\left(2^2\cdot3\right)^3\cdot\left(2^3\right)^3\cdot5}{2^9\cdot3^2\cdot5^2}\)
A=\(\dfrac{\left(2^2\right)^3\cdot3^3\cdot2^9\cdot5}{2^9\cdot3^2\cdot5^2}\)
A=\(\dfrac{2^6\cdot3^3\cdot2^9\cdot5}{2^9\cdot3^2\cdot5^2}\)
A=\(\dfrac{2^6\cdot3}{5}\)
A=\(\dfrac{64\cdot3}{5}\)
A=\(\dfrac{192}{5}\)
Vậy A=\(\dfrac{192}{5}\)
Tích cho mình nhé!
LB
1
22 tháng 9 2017
\(A=\dfrac{\left(12.8\right)^3.5}{2^9.15^2}=\dfrac{\left(2^2.3.2^3\right)^3.5}{2^9.\left(3.5\right)^2}=\dfrac{2^6.3^3.2^9.5}{2^9.3^2.5^2}=\dfrac{2^{15}.3^3.5}{2^9.3^2.5^2}=\dfrac{2^6.3.1}{1.1.5}=\dfrac{192}{5}\)
N
19 tháng 7 2017
\(A=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)
\(A=\frac{2.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}\)
\(A=\frac{2}{3}\)
NH
1
NA
1
17 tháng 2 2020
Ta có :\(B=\frac{3}{2}-\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3-....-\left(\frac{3}{2}\right)^{2014}\)
\(\frac{3}{2}B=\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4-...+\left(\frac{3}{2}\right)^{2014}-\left(\frac{3}{2}\right)^{2015}\)
\(\frac{3}{2}B+B=\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^3+..+\left(\frac{3}{2}\right)^{2014}-\left(\frac{3}{2}\right)^{2015}\) \(+\frac{3}{2}-\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3-...-\left(\frac{3}{2}\right)^{2014}\)
\(\frac{5}{2}B=\frac{3}{2}-\left(\frac{3}{2}\right)^{2015}\)
\(B=\frac{\frac{3}{2}-\left(\frac{3}{2}\right)^{2015}}{\frac{5}{2}}\)
