CMR:
51/2 . 52/2 . 53/2 . ... . 100/2 = 1 . 3 . 5 . ... . 99
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L
1
11 tháng 4 2023
Sửa đề: \(\dfrac{\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)
=1
NA
2
PT
0
D
2
20 tháng 3 2017
Đặt \(A=1.3.5.7...99\)
\(B=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}\)
Ta có:
\(A=1.3.5.7...99\)
\(\Rightarrow A=\dfrac{\left(1.3.5.7...99\right)\left(2.4.6.8...100\right)}{2.4.6.8...100}\)
\(\Rightarrow A=\dfrac{1.2.3.4...100}{2.4.6.8...100}\)
\(\Rightarrow A=\dfrac{1.2.3.4...100}{\left(2.1\right)\left(2.2\right)\left(2.3\right)...\left(2.50\right)}\)
\(\Rightarrow A=\dfrac{\left(1.2.3.4...50\right)\left(51.52.53...100\right)}{\left(1.2.3.4...50\right)\left(2.2.2.2...2\right)}\)
\(\Rightarrow A=\dfrac{51.52.53.54...100}{2.2.2.2...2}\)
\(\Rightarrow A=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}....\dfrac{100}{2}\)
\(\Rightarrow A=B\)
Vậy \(1.3.5.7...99=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}\) (Đpcm)
20 tháng 3 2017
VT: 1.3.5.7....99=\(\dfrac{(1.3.5.7.....99).\left(2.4.6....100\right)}{2.4.6....100}\)
\(=\dfrac{\left(1.3.5.7.....99\right)\left(2.4.6.....100\right)}{1.2.2.2.2.3.....2.50}\)\(=\dfrac{\left(1.2.3.4.....50\right)\left(51.52.53....100\right)}{\left(1.2.3.4......50\right)\left(2.2.2.2.2....2\right)}\)
\(=\dfrac{51.52.53......100}{2.2.2.2.....2}=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}......\dfrac{100}{2}=VP\left(đpcm\right)\)
NV
0
NM
1
Đặt C=1.3.5.7...99
Đặt D=51/2.52/2.53/2 ....100/2
Ta có:C=1.3.5.7...99
=>2.4.6...100.C=1.2.3...100
=>C = (1.2.3....100) / (2.4.6...100)= (1.2.3...50).(51.52...100) / [(2.1)(2.2).(2.3)...(2.50)]
C=(1.2.3...50).(51.52...100) /[2^50.(1.2.3...50)] =(51.52...100)/2^50 =51/2.52/2.53/2...100/2 =D
Vậy C=D
Ta có :
\(1.3.5.....99=\frac{\left(1.3.5.....99\right)\left(2.4.6.....98\right)}{2.4.6.....98}=\frac{1.2.3.....99.100}{2^{50}\left(1.2.3.....50\right)}=\frac{51.52.53.....100}{2.2.2.....2}\)
\(=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}.....\frac{100}{2}\)
Vậy......................
~ Hok tốt ~