So sánh:
A= \(\dfrac{1.3.5+2.6.10+4.12.20}{1.5.7+2.10.12+4.20.28}\)với\(\dfrac{3}{8}\)
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HK
1
9 tháng 8 2018
\(\frac{1.3.5+2.6.10+4.12.20}{1.5.7+2.10.14+4.20.28}\)
\(=\frac{3.5+2.3.2.5.2+4.3.4.5.4}{5.7+2.5.2.2.7+4.4.5.7.4}\)
\(=\frac{3.5.\left(1+2.2.2+4.4.4\right)}{5.7.\left(1+2.2.2+4.4.4\right)}\)
\(=\frac{3}{7}>\frac{3}{8}\)
LT
0
NT
2
2 tháng 7 2017
\(\dfrac{1.3.5+2.6.10+4.12.20+7.21.35}{1.5.7+2.10.14+4.20.28+7.35.49}\)
\(=\dfrac{1.3.5+2^3.1.3.5+2^6.1.3.5+7^3.1.3.5}{1.5.7+2^3.1.5.7+2^6.1.5.7+7^3.1.5.7}\)
\(=\dfrac{1.3.5\left(1+2^3+2^6+7^3\right)}{1.5.7\left(1+2^3+2^6+7^3\right)}\)
\(=\dfrac{1.3.5}{1.5.7}\)
\(=\dfrac{3}{7}\)
2 tháng 7 2017
Ta có : \(\dfrac{1.3.5+2.6.10+4.12.20 +7.21.35 }{1.5.7+2.10.14+4.20.28+7.35.49}\)
\(=\dfrac{1.3.5+1.2.3.2.5.2+1.4.3.4.5.4+1.7.3.7.5.7}{1.5.7+1.2.5.2.7.2+1.4.5.4.7.4+1.7.5.7.7.7}\)
\(=\dfrac{1.\left(1.3.5\right)+2.\left(1.3.5\right)+4.\left(1.3.5\right)+7.\left(1.3.5\right)}{1.\left(1.5.7\right)+2.\left(1.5.7\right)+4.\left(1.5.7\right)+7.\left(1.5.7\right)}\)
\(=\dfrac{1.3.5.\left(1+2+4+7\right)}{1.5.7.\left(1+2+4+7\right)}\)
\(=\dfrac{3}{7}\)
NQ
0
NT
1
2 tháng 7 2017
\(\frac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot49}\)
=\(\frac{3\cdot\left(1\cdot5+2\cdot2\cdot10+4\cdot4\cdot20+7\cdot7\cdot35\right)}{7\cdot\left(1\cdot5+2\cdot10\cdot2+4\cdot20\cdot4+35\cdot49\right)}\)=\(\frac{3}{7}\)
PV
1
FL
0
\(A=\dfrac{15\left(1+2\cdot4+64\right)}{35+240+2240}\)
\(=\dfrac{15\cdot73}{2515}=\dfrac{15\cdot73}{5\cdot503}=\dfrac{3\cdot73}{503}=\dfrac{219}{503}>\dfrac{3}{8}\)