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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 \(A=B\)
Ta có:
\(1.2.3....99=\dfrac{1.2.3.4....99.100}{2.4.6....100}\)
\(=\dfrac{1}{2.1}.\dfrac{2}{2.2}...\dfrac{100}{2.50}\)
\(=\dfrac{1.2.3.4....100}{1.2.3....50.2.2.2...2}\) ( \(50\) thừa số \(2\) )
\(=\dfrac{51.51...100}{2.2.2...2}\)
\(=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}\)
\(\Rightarrow A=B\left(đpcm\right)\)
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=\(\frac{51}{2}\cdot\frac{52}{2}\cdot\frac{53}{2}\cdot...\cdot\frac{100}{2}=B\)
Lời giải:
\(A=1.3.5.7...99=\frac{1.2.3.4...99.100}{2.4.6.8.100}=\frac{1.2.3...99.100}{(1.2)(2.2)(3.2)...(50.2)}\)
\(=\frac{1.2.3...99.100}{(1.2.3...50).2^{50}}=\frac{51.52...100}{2^{50}}=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}=B\)
\(A=1.3.5.7...99=\frac{\left(1.3.5.7...99\right)\left(2.4.6...100\right)}{2.4.6...100}=\frac{1.2.3...100}{\left(2.1\right)\left(2.2\right)...\left(2.50\right)}=\frac{\left(1.2.3...50\right)\left(51.52.53....100\right)}{\left(1.2.3...50\right)\left(2.2.2...2\right)}=\frac{51.52.53...100}{2.2...2}=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}=B\)
\(1.3.....99=\frac{1.3....99.2.4.6....100}{2.4.6....100}\)
\(=\frac{1.2.3.4.5......99.100}{2^{50}.\left(1.2.3....50\right)}\)
\(=\frac{51.52.53...100}{2.2.2...2}\)
\(=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}\)
\(\Rightarrow1.3...99=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}\left(đpcm\right)\)
Ta có :\(\frac{51}{2}\) . \(\frac{52}{2}\) .... \(\frac{100}{2}\)
=\(\frac{51.52....100}{2.2....2}\)
=\(\frac{51.52....100}{2.2....2}\) . \(\frac{2.4.6....100}{2.4.6....100}\)
=\(\frac{51.52....100.2.4.6...100}{2.4.6...100.2.2...2}\)
=\(\frac{1.2.3.4...100}{2.4.6...100}\)
=\(\frac{\left[1.3.5....99\right].\left[2.4.6...100\right]}{2.4.6...100}\)
=1.3.5...99[đpcm]
bang nhau
Giai:
A=1.3.5.7...97.99=\(\frac{\left(1.3.5...97.99\right).\left(2.4.6...100\right)}{2.4.6...100}\)
=\(\frac{1.2.3.4...99.100}{\left(1.2\right).\left(2.2\right)...\left(2.50\right)}\)
=\(\frac{\left(1.2.3...50\right).\left(51.52...99.100\right)}{\left(1.2.3...49.50\right).2^{50}}\)
=\(\frac{51.52...99.100}{2.2...2.2}\)
=\(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}\)
mà B=\(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}\)
Nên A=B
Vậy A=B
\(1.3.5.7...97.99=\frac{100!}{2.4.6.8...100}\)
\(=\frac{1.2.3.4...100}{1.2.2.2.3.2...50.2}\)
\(=\frac{51.52.53...100}{2}\)
Vậy \(A=B\)
Ta có:
\(B=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}=\dfrac{51.52.53...100}{2^{50}}\)
\(=\dfrac{\left(51.52.53..100\right)\left(1.2.3.4...50\right)}{2^{50}\left(1.2.3.4...50\right)}\)
\(=\dfrac{1.2.3.4.5.6...100}{\left(2.1\right)\left(2.2\right)\left(2.3\right)...\left(2.50\right)}\)
\(=\dfrac{1.2.3.4.5.6...100}{2.4.6.8.10...100}=\dfrac{\left(1.3...99\right)\left(2.4...100\right)}{2.4.6...100}\)
\(=1.3.5.7.99=A\)
Vậy \(A=B\) (Đpcm)