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\(C=\frac{\left(1.3.5.7...99\right).\left(2.4.6.8...98.100\right)}{2.4.6.8....98.100}=\frac{\left(1.2.3.4...99.100\right)}{\left(1.2\right).\left(2.2\right)\left(3.2\right)...\left(49.2\right).\left(50.2\right)}=\frac{1.2.3.4....99.100}{\left(1.2.3...49.50\right).\left(2.2.2...2\right)}\)
\(C=\frac{\left(1.2.3.4...50\right).\left(51.52.53...100\right)}{\left(1.2.3...50\right)\left(2.2.2....2\right)}=\frac{51}{2}.\frac{52}{2}...\frac{100}{2}=D\)
Vậy C = D
a, 2 + 4 + 6 +...+ 2x = 210
=> 2(1 + 2 + 3 +...+ x) = 210
=> \(\frac{2x\left(x+1\right)}{2}=210\)
=> x(x + 1) = 210
=> x(x + 1) = 14.15
=> x = 14
b, Ta có: \(B=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}....\frac{100}{2}=\frac{51.52.53....100}{2^{50}}\)
\(=\frac{\left(51.52.53....100\right)\left(1.2.3.....50\right)}{2^{50}\left(1.2.3.....50\right)}\)
\(=\frac{1.2.3.....100}{\left(2.1\right)\left(2.2\right)\left(2.3\right)....\left(2.50\right)}\)
\(=\frac{\left(1.3.5....99\right)\left(2.4.6....100\right)}{2.4.6.....100}\)
\(=1.3.5.....99=B\)
Vậy A = B
