\(\frac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot49}\)
=\(\frac{3\cdot\left(1\cdot5+2\cdot2\cdot10+4\cdot4\cdot20+7\cdot7\cdot35\right)}{7\cdot\left(1\cdot5+2\cdot10\cdot2+4\cdot20\cdot4+35\cdot49\right)}\)=\(\frac{3}{7}\)

\(\dfrac{1.3.5+2.6.10+4.12.20+7.21.35}{1.5.7+2.10.14+4.20.28+7.35.49}\)

\(=\dfrac{1.3.5+2^3.1.3.5+2^6.1.3.5+7^3.1.3.5}{1.5.7+2^3.1.5.7+2^6.1.5.7+7^3.1.5.7}\)

\(=\dfrac{1.3.5\left(1+2^3+2^6+7^3\right)}{1.5.7\left(1+2^3+2^6+7^3\right)}\)

\(=\dfrac{1.3.5}{1.5.7}\)

\(=\dfrac{3}{7}\)

Ta có : \(\dfrac{1.3.5+2.6.10+4.12.20 +7.21.35 }{1.5.7+2.10.14+4.20.28+7.35.49}\)

\(=\dfrac{1.3.5+1.2.3.2.5.2+1.4.3.4.5.4+1.7.3.7.5.7}{1.5.7+1.2.5.2.7.2+1.4.5.4.7.4+1.7.5.7.7.7}\)

\(=\dfrac{1.\left(1.3.5\right)+2.\left(1.3.5\right)+4.\left(1.3.5\right)+7.\left(1.3.5\right)}{1.\left(1.5.7\right)+2.\left(1.5.7\right)+4.\left(1.5.7\right)+7.\left(1.5.7\right)}\)

\(=\dfrac{1.3.5.\left(1+2+4+7\right)}{1.5.7.\left(1+2+4+7\right)}\)

\(=\dfrac{3}{7}\)

\(A=\dfrac{15\left(1+2\cdot4+64\right)}{35+240+2240}\)

\(=\dfrac{15\cdot73}{2515}=\dfrac{15\cdot73}{5\cdot503}=\dfrac{3\cdot73}{503}=\dfrac{219}{503}>\dfrac{3}{8}\)

Xét số hạng tổng quát: 
1 + 1/[k.(k + 2)] = [k.(k + 2) + 1]/[k.(k + 2)] = (k + 1)²/[k.(k + 1)], với k nguyên dương. 
Cho k chạy từ 1 đến 99, ta có: 
• 1 + 1/1.3 = 2²/(1.3). 
• 1 + 1/2.4 = 3²/(2.4). 
• 1 + 1/3.5 = 4²/(3.5). 
....................... 
• 1 + 1/97.99 = 98²/(97.99). 
• 1 + 1/98.100 = 99²/(98.100). 
• 1 + 1/99.101 = 100²/(99.101). 
Nhân vế với vế các đẳng thức trên, ta được: 
(1 + 1/1.3).(1 + 1/2.4)(1 + 1/3.5)....(1 + 1/99.101) 
= [2².3².....100²]/[1.2.3².4²......99².100...‡ 
= (2².100²)/(2.100.101) 
= 2.100/101 
= 200/101.

còn N thì chịu

M=(4/1.3.9/2.4.16/3.5...10000/99.101

M=2.2/1.3.3.3/2.4.4.4/3.5...100.100/99.101

M=2.3.4.5...100/1.2.3...99.3.4.5...100/2.3.4.5...101

M=100.2/101=200/101

Cau N sai de rui ban a, o mau so phai la 1.5.7+2.10.14+4.20.28+7.35.49 moi lam dc.

mấy bn giải khó hiểu wa .....................