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\(B=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}.....\frac{100}{2}\)
\(B=\frac{51.52.53...100}{2.2.2.2.....2}=\frac{51.52.53....100}{2^{50}}=\frac{\left(1.2.3.4....50\right).\left(51.52.53...100\right)}{\left(1.2.3....50\right).2^{50}}\)
\(B=\frac{1.2.3.4.5.....98.99.100}{\left(1.2\right).\left(2.2\right).\left(2.3\right)....\left(2.50\right)}=\frac{1.2.3.4.5....98.99.100}{2.4.6......100}\)
\(B=1.3.5....99=A\)
Vậy \(A=B\)
Ta có :
\(A=1.3.5.7...99\)
\(A=\frac{\left(1.3.5.7...99\right).\left(2.4.6...100\right)}{2.4.6...100}\)
\(A=\frac{1.2.3.4.5.6.7...99.100}{\left(2.2...2\right).\left(1.2.3...50\right)}\)
\(A=\frac{\left(1.2.3...50\right).\left(51.52...100\right)}{2^{50}.\left(1.2.3...50\right)}\)
\(A=\frac{51.52...100}{2^{50}}\)
Mà \(B=\frac{51}{2}.\frac{52}{2}...\frac{100}{2}\)\(=\frac{51.52...100}{2^{50}}\)
vậy \(A=B\)
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
VAy C=D
\(A=\frac{1.2.3...........99.100}{2.4.6....100}\)
\(=\frac{1.2.3..............99.100}{1.2.2.2.2.3.........50.2}\)
\(=\frac{1.2.3.......50........99.100}{\left(1.2.3........50\right).2.2.....2}\)
\(=\frac{51.52..........99.100}{2.2............2}\)
\(=\frac{51}{2}.\frac{52}{2}...........\frac{100}{2}\)
Ta có :A= (1*3*5*7*...*99)*(2*4*6*...*100):(2*4*6*..*100)
A=\(\frac{1\cdot2\cdot3\cdot4\cdot...\cdot100}{2\cdot4\cdot6\cdot...\cdot100}=\frac{\left(1\cdot2\cdot3\cdot4\cdot...\cdot50\right)\cdot\left(51\cdot52\cdot53\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot50\right)\cdot\left(2\cdot2\cdot2\cdot...\cdot2\right)}\)(MẤU TÁCH 2 RA NGOÀI)
A=\(\frac{51\cdot52\cdot53\cdot...\cdot100}{2\cdot2\cdot2\cdot..\cdot2}\)
A=\(\frac{51}{2}\cdot\frac{52}{2}\cdot\frac{53}{2}\cdot...\cdot\frac{100}{2}=B\)