Lời giải:

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

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

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 ~

cách viết phân số trên máy tính là j v mn

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

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

- Ta có : `C=51/2 * 52/2 * 53/2* ... * 100/2`

`-> C=(51.52.53...100)/(2^50)`

`-> C=((1.2.3...50).(51.52.53...100))/((1.2.3...50).2^50)`

`-> C=(1.2.3...100)/((1.2).(2.2).(3.2)...(50.2))`

`-> C=(1.2.3...100)/(2.4.6...100)`

`-> C=1.3.5.7...99`

- Từ đó ta có :

`B-C=1.3.5.7...99-1.3.5.7...99=0`

- Vậy `B-C=0`

Ta có:

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

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

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

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

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

\(\Rightarrow A=B\)

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