Lời giải:

\(A=1.3.5.7...99=\frac{1.2.3.4...99.100}{2.4.6.8.100}=\frac{1.2.3...99.100}{(1.2)(2.2)(3.2)...(50.2)}\)

\(=\frac{1.2.3...99.100}{(1.2.3...50).2^{50}}=\frac{51.52...100}{2^{50}}=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}=B\)

\(1\cdot3\cdot5\cdot...\cdot99=\dfrac{\left(1\cdot3\cdot5\cdot...\cdot99\right)\cdot\left(2\cdot4\cdot6\cdot...\cdot100\right)}{2\cdot4\cdot6\cdot...\cdot100}\)

\(=\dfrac{1\cdot3\cdot5\cdot...\cdot2\cdot4\cdot6\cdot...\cdot100}{1\cdot2\cdot3\cdot...\cdot50\cdot2\cdot2\cdot...\cdot2}=\dfrac{51}{2}\cdot\dfrac{52}{2}\cdot...\cdot\dfrac{100}{2}\)

Ta có:

\(\dfrac{51}{2}\cdot\dfrac{52}{2}\cdot...\cdot\dfrac{100}{2}\\ =\dfrac{51\cdot52\cdot...\cdot100}{2^{50}}\\ =\dfrac{\left(1\cdot2\cdot...\cdot50\right)\left(51\cdot52\cdot...\cdot100\right)}{\left(1\cdot2\cdot...\cdot50\right)\cdot2^{50}}\\ =\dfrac{1\cdot2\cdot3\cdot...\cdot100}{2\cdot4\cdot6\cdot...\cdot100}\\ =1\cdot3\cdot5\cdot...\cdot99\)

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)

VT: 1.3.5.7....99=\(\dfrac{(1.3.5.7.....99).\left(2.4.6....100\right)}{2.4.6....100}\)

\(=\dfrac{\left(1.3.5.7.....99\right)\left(2.4.6.....100\right)}{1.2.2.2.2.3.....2.50}\)\(=\dfrac{\left(1.2.3.4.....50\right)\left(51.52.53....100\right)}{\left(1.2.3.4......50\right)\left(2.2.2.2.2....2\right)}\)

\(=\dfrac{51.52.53......100}{2.2.2.2.....2}=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}......\dfrac{100}{2}=VP\left(đpcm\right)\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-2\cdot\dfrac{1}{2}-2\cdot\dfrac{1}{4}-...-2\cdot\dfrac{1}{100}\)

\(A=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-\dfrac{1}{1}-\dfrac{1}{2}-...-\dfrac{1}{50}\)

\(A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)

\(\Rightarrow A=B\)

tớ giải chi tiết hơn nhá:

A=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

A=(\(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{99}-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)

A=\(\left(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)

A=\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)

Vậy A=B

b)Ta có :

\(A=1.3.5...........99\)

\(\Rightarrow A=\dfrac{\left(1.3.7.9.............99\right)\left(2.4.6.8........100\right)}{2.4.6.8.............100}\)

\(\Rightarrow A=\dfrac{1.2.3.4.............100}{2.4.6.8................100}\)

\(\Rightarrow A=\dfrac{1.2.3.4..................100}{\left(2.1\right)\left(2.2\right)...............\left(2.50\right)}\)

\(\Rightarrow A=\dfrac{51.52.53...........................100}{2.2.2.2.............................2}\)

\(\Rightarrow A=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}.............\dfrac{100}{2}\)

\(\Rightarrow A=D\)

~ Chúc bn học tốt ~

mk hỏi này sao mà 1.2.3.4.....100/(2.1).(2.2)...(2.50)lại =51.52.53..........100/2.2.2........2