\(A=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\frac{1}{99\cdot101}\right)\)

\(A=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot...\cdot\frac{10000}{99\cdot101}\)

\(A=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)\cdot...\cdot\left(100\cdot100\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(99\cdot101\right)}\)

\(A=\frac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot...\cdot101\right)}\)

\(A=\frac{100\cdot2}{1\cdot101}\)

\(A=\frac{200}{101}\)

Đáp án C

Với B = 3

Ta có: 

a) 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/x.(x+1) = 499/500

1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/x - 1/x+1 = 499/500

1 - 1/x+1 = 499/500

1/x+1 = 1 - 499/500

1/x+1 = 1/500

x + 1 = 500

     x = 500 - 1 

     x = 499

b) 1/1.3 + 1/3.5 + 1/5.7 + .... + 1/x.(x+2) = 20/41

1/2 . [ 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/x.(x+2) ] = 20/41

1/2 . [ 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x+2 ] = 20/41

1/2 . [ 1 - 1/x+2 ) = 20/41

1 - 1/x+2 = 20/41 : 1/2

1 - 1/x+2 = 40/41

1/x+2 = 1 - 40/41

1/x+2 = 1/41

x + 2 = 41

      x = 41 - 2 

      x = 39

a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)

Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)

\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)

\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)

\(\dfrac{1}{x+1}=\dfrac{1}{100}\)

\(\Rightarrow x+1=100\)

\(x=100-1\)

\(x=99\)

câu b thiếu kết quả đúng không bn?

Công thức: \(\dfrac{1}{a\times b}=\) 1/ khoảng cách giữa a và b \(\times\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\)

* Bạn làm theo công thức và vẫn dụng câu b nhé.

ai nhanh mk k

trả lời em dùm câu hỏi phía dưới vs

mk cx ko bt mà mk đag cần gấp

\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)y=\frac{2}{3}\)

=> \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)

=> \(\frac{1}{2}\left(1-\frac{1}{11}\right)y=\frac{2}{3}\)

=> \(\frac{1}{2}.\frac{10}{11}y=\frac{2}{3}\)

=> \(\frac{5}{11}y=\frac{2}{3}\)

=>y = \(\frac{2}{3}:\frac{5}{11}\)

=> y = \(\frac{22}{15}\)

cho mk cái lời giải thích chỗ nhân 1/2 ý mk ko hiểu mong bn thông cảm

bạn phạm khánh hà ơi dấu chấm ở giữa các phân số có nghĩa là dấu nhân đó

\(\left(x+1\right)^6=C^k_6.x^k\)

\(x^5\Rightarrow k=5\Rightarrow C^5_6\)

Tuong tu: \(C^5_7;C^5_8;C^5_9;C^5_{10};C^5_{11};C^5_{12}\)

\(\Rightarrow he-so:C^5_7+C^5_8+...+C^5_{12}=...\)