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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)
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=\(\frac{51}{2}\cdot\frac{52}{2}\cdot\frac{53}{2}\cdot...\cdot\frac{100}{2}=B\)
\(B=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}.....\frac{100}{2}\)
\(B=\frac{51.52.53...100}{2.2.2.2.....2}=\frac{51.52.53....100}{2^{50}}=\frac{\left(1.2.3.4....50\right).\left(51.52.53...100\right)}{\left(1.2.3....50\right).2^{50}}\)
\(B=\frac{1.2.3.4.5.....98.99.100}{\left(1.2\right).\left(2.2\right).\left(2.3\right)....\left(2.50\right)}=\frac{1.2.3.4.5....98.99.100}{2.4.6......100}\)
\(B=1.3.5....99=A\)
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 \(A=B\)
a) \(\frac{2}{3}x\times\frac{1}{2}=\frac{1}{10}\Rightarrow\frac{2}{3}x=\frac{1}{5}\Rightarrow x=\frac{3}{10}\)
Đặt A=\(\frac{1}{3}.5+\frac{1}{5}.7+...+\frac{1}{97}.99\)
=>A=\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
=>2A=\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
=>2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
=>2A=\(\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
=>A=\(\frac{32}{99}:2=\frac{32}{99}.\frac{1}{2}=\frac{32}{198}=\frac{16}{99}\)
\(A=\frac{1.2.3...........99.100}{2.4.6....100}\)
\(=\frac{1.2.3..............99.100}{1.2.2.2.2.3.........50.2}\)
\(=\frac{1.2.3.......50........99.100}{\left(1.2.3........50\right).2.2.....2}\)
\(=\frac{51.52..........99.100}{2.2............2}\)
\(=\frac{51}{2}.\frac{52}{2}...........\frac{100}{2}\)
Ta có:
\(1.2.3....99=\dfrac{1.2.3.4....99.100}{2.4.6....100}\)
\(=\dfrac{1}{2.1}.\dfrac{2}{2.2}...\dfrac{100}{2.50}\)
\(=\dfrac{1.2.3.4....100}{1.2.3....50.2.2.2...2}\) ( \(50\) thừa số \(2\) )
\(=\dfrac{51.51...100}{2.2.2...2}\)
\(=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}\)
\(\Rightarrow A=B\left(đpcm\right)\)