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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1.3.5+2.6.10+4.12.20+7.21.35}{1.5.7+2.10.14+4.20.28+7.35.49}\)
\(=\dfrac{1.3.5+2^3.1.3.5+2^6.1.3.5+7^3.1.3.5}{1.5.7+2^3.1.5.7+2^6.1.5.7+7^3.1.5.7}\)
\(=\dfrac{1.3.5\left(1+2^3+2^6+7^3\right)}{1.5.7\left(1+2^3+2^6+7^3\right)}\)
\(=\dfrac{1.3.5}{1.5.7}\)
\(=\dfrac{3}{7}\)
Ta có : \(\dfrac{1.3.5+2.6.10+4.12.20 +7.21.35 }{1.5.7+2.10.14+4.20.28+7.35.49}\)
\(=\dfrac{1.3.5+1.2.3.2.5.2+1.4.3.4.5.4+1.7.3.7.5.7}{1.5.7+1.2.5.2.7.2+1.4.5.4.7.4+1.7.5.7.7.7}\)
\(=\dfrac{1.\left(1.3.5\right)+2.\left(1.3.5\right)+4.\left(1.3.5\right)+7.\left(1.3.5\right)}{1.\left(1.5.7\right)+2.\left(1.5.7\right)+4.\left(1.5.7\right)+7.\left(1.5.7\right)}\)
\(=\dfrac{1.3.5.\left(1+2+4+7\right)}{1.5.7.\left(1+2+4+7\right)}\)
\(=\dfrac{3}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1.2.3+2.4,6+4.8.12+7.14.21}{1.3.5+2.6.10+4.12.20+7.21.35}\)
\(=\frac{1\left(1.2.3\right)+2\left(1.2.3\right)+4\left(1.2.3\right)+7\left(1.2.3\right)}{1\left(1.3.5\right)+2\left(1.3.5\right)+4\left(1.2.3\right)+7\left(1.2.3\right)}\)
\(=\frac{6\left(1+2+4+7\right)}{15\left(1+2+4+7\right)}=\frac{6}{15}=\frac{3}{5}\)
tính hợp lý \(\frac{9764}{36615}+\frac{36\cdot85\cdot20}{25\cdot84\cdot34}+2\cdot2+3\frac{19}{133}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
C=\(\frac{1.2.3+2.4.6+4.8.12+7.14.21}{1.3.5+2.6.10+4.12.20+7.21.35}+\frac{74.147-73}{73.147+74}+216,6\)
=\(\frac{2+4+14}{5+10+35}+0+216,6\)
=\(\frac{2+2+2}{5+5+5}+0+216,6\)
=0+0+216,6
=216,6