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NN
2
20 tháng 4 2017
A=5.(2.3+4.9.5+6.9.7+10.21.8)/7.(2.3+4.9.5+6.9.7+10.21.8)
A=5/7
N
1
LP
0
DX
2
SI
7 tháng 3 2021
A = \(\dfrac{2.3.5+4.9.35+6.9.35+10.21.40}{2.3.7+4.9.35+6.9.49+10.21.56}\)
A = \(\dfrac{2.3.5+\left(2.3.5\right).2.3.5+\left(2.3.5\right).3.3.7+\left(2.3.5\right).5.7.8}{2.3.7+\left(2.3.7\right).2.3.5+\left(2.3.7\right).3.3.7+\left(2.3.7\right).5.7.8}\)
A = \(\dfrac{2.3.5.\left(1+2.3.5+3.3.7+5.7.8\right)}{2.3.7\left(1+2.3.5+3.3.7+5.7.8\right)}\)
A = \(\dfrac{2.3.5}{2.3.7}\)
A = \(\dfrac{5}{7}\)
7 tháng 3 2021
Ta có: \(A=\dfrac{2\cdot3\cdot5+4\cdot9\cdot35+6\cdot9\cdot35+10\cdot21\cdot40}{2\cdot3\cdot7+4\cdot9\cdot35+6\cdot9\cdot49+10\cdot21\cdot56}\)
\(=\dfrac{2\cdot3\cdot5+2\cdot3\cdot5\cdot6\cdot7+2\cdot3\cdot5\cdot3\cdot3\cdot7+2\cdot3\cdot5\cdot5\cdot7\cdot8}{2\cdot3\cdot7+2\cdot3\cdot7\cdot2\cdot3\cdot5+2\cdot3\cdot7\cdot3\cdot3\cdot7+2\cdot3\cdot7\cdot5\cdot7\cdot8}\)
\(=\dfrac{2\cdot3\cdot5\cdot\left(1+2\cdot3\cdot7+3\cdot3\cdot7+5\cdot7\cdot8\right)}{2\cdot3\cdot7\cdot\left(1+2\cdot3\cdot5+3\cdot3\cdot7+5\cdot7\cdot8\right)}\)
\(=\dfrac{5}{7}\cdot\dfrac{386}{374}=\dfrac{965}{1309}\)
LP
0
TA
0
U
21 tháng 8 2018
\(B=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{400}-1\right)\)
\(-B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{400}\right)\)
\(-B=\frac{3}{4}\cdot\frac{5}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{399}{400}\)
\(-B=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(19\cdot21\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(20\cdot20\right)}\)
\(-B=\frac{\left(1\cdot2\cdot3\cdot...\cdot19\right)\left(3\cdot4\cdot5\cdot...\cdot21\right)}{\left(2\cdot3\cdot4\cdot...\cdot20\right)\left(2\cdot3\cdot4\cdot...\cdot20\right)}\)
\(-B=\frac{1\cdot21}{20\cdot2}\)
\(-B=\frac{21}{40}\)
\(B=\frac{-21}{40}\)
21 tháng 8 2018
\(A=\frac{\text{2.3.5+2.2.3.3.5.5+2.5.3.7.2.2.2.5 }}{2.3.7+2.2.3.3.7.5+2.3.3.3.7.7+2.5.3.7.7.2.2.2}\)
\(=\frac{2.3.5.\left(1+2.3.5+2^3.5.7\right)}{2.3.7.\left(1+2.3.5+3.3.7+2^3.5.7\right)}\)
\(=\frac{2.3.5}{2.3.7.3.3.7}=\frac{5}{441}\)
mk mới học lớp 5
Ta có : \(A=\frac{2.3.5.1.1.1+2.3.5.2.3.5+2.3.5.3.3.7+2.3.5.5.7.8}{2.3.7.1.1.1+2.3.7.2.3.5+2.3.7.3.3.7+2.3.7.5.7.8}\)
\(=\frac{2.3.5\left(1.1.1+2.3.5+3.3.7+5.7.8\right)}{2.3.7\left(1.1.1+2.3.5+3.3.7+5.7.8\right)}=\frac{2.3.5}{2.3.7}=\frac{5}{7}\)