\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+...+\frac{1}{97\cdot3}+\frac{1}{99\cdot1}}=\frac{\left[1+\frac{1}{99}\right]+\left[\frac{1}{3}+\frac{1}{97}\right]+...+\left[\frac{1}{49}+\frac{1}{51}\right]}{2\left[\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+...+\frac{1}{49\cdot51}\right]}\)

\(=\frac{\frac{100}{1\cdot99}+\frac{100}{3\cdot97}+...+\frac{100}{49.51}}{2\left[\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+...+\frac{1}{49.51}\right]}=\frac{100\left[\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+...+\frac{1}{49.51}\right]}{2\left[\frac{1}{1\cdot99}+\frac{1}{3\cdot97}+...+\frac{1}{49.51}\right]}=\frac{100}{2}=50\)

Tử số = 1 + 1/3 + 1/5 + ... + 1/97 + 1/99

= (1 + 1/99) + (1/3 + 1/97) + ... + (1/49 + 1/51)

= 100/1.99 + 100/3.97 + ... + 100/49.51

= 100.(1/1.99 + 1/3.97 + ... + 1/49.51)

Mẫu số = 1/1.99 + 1/3.97 + 1/5.95 + ... + 1/97.3 + 1/99.1

= 2.(1/1.99 + 1/3.97 + 1/5.95 + ... + 1/49.51)

=> phân số đề bài cho = 100/2 = 50

                    Ta có :

               \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}}\)

              \(=\frac{\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)}{2.\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}\right)}\)

             \(=\frac{\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}}{2.\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)

             \(=\frac{100.\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}{2.\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}\)

             \(=\frac{100}{2}=50\)

            Ủng hộ mk nha !!! ^_^

Tính B=(1/2+1/3+1/4+....+1/100)/(99/1+98/2+97/3+...+1/99)

nhân 100 với cả tử với mẫu sau đó phân tích mẫu ta có

100/1.99 +100/3.97+ ...+100/99.1 = (1+99)/1.99 +(3+97)/3.97 +...+ (1+99)/1.99

= 1/1.99+99/1.99+3/3.97+97/3.97+...+1/1.99+99/1.99

= 1/99 + 1/1 +1/3 +1/97+....+1/97+1/3+1/99+1/1

= 2.(1+1/3+1/5+1/7+.....1/97+1/99)

Do đó A =[100(1+1/3+1/5+..+1/99)] / 2(1+1/3+1/5+...1/99) = 50

nhân 100 với cả tử với mẫu sau đó phân tích mẫu ta có

100/1.99 +100/3.97+ ...+100/99.1 = (1+99)/1.99 +(3+97)/3.97 +...+ (1+99)/1.99

= 1/1.99+99/1.99+3/3.97+97/3.97+...+1/1.99+99/1.99

= 1/99 + 1/1 +1/3 +1/97+....+1/97+1/3+1/99+1/1

= 2.(1+1/3+1/5+1/7+.....1/97+1/99)

Do đó A =[100(1+1/3+1/5+..+1/99)] / 2(1+1/3+1/5+...1/99) = 50

\(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{99}\)

\(=\left(\frac{1}{99}+1\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+......+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\left(\frac{100}{1.99}\right)+\left(\frac{100}{3.97}\right)+.....+\left(\frac{100}{49.51}\right)\)

\(Ta\)\(thấy\)\(Biểu\)\(Thức\)\(Này\)\(gấp\)\(50\)\(biểu\)\(thức\)\(kia\)

\(\Rightarrow A=50\)

nếu biết tách mẫu thì mẫu sẽ gấp 100 lần tử nhé

à anh xin lỗi kết quả phải là 50

Ta có:\(\frac{1}{1.99}+\frac{1}{3.97}+.............+\frac{1}{99.1}\)

=\(\frac{1}{100}.\left(\frac{1}{1}+\frac{1}{99}+\frac{1}{3}+\frac{1}{97}+.........+\frac{1}{99}+\frac{1}{1}\right)\)

=\(\frac{1}{50}.\left(\frac{1}{1}+\frac{1}{3}+........+\frac{1}{97}+\frac{1}{99}\right)\)

Vậy A=50

Q=\(\frac{3+1+\frac{3}{5}+...+\frac{3}{99}}{\left(\frac{1}{1.99}+\frac{1}{99.1}\right)+\left(\frac{1}{3.97}+\frac{1}{97.3}\right)+...+\left(\frac{1}{49.51}+\frac{1}{51.49}\right)}\)

Q=\(\frac{\frac{3}{1}+\frac{3}{3}+\frac{3}{5}+...+\frac{3}{99}}{\frac{2}{1.99}+\frac{2}{3.97}+...+\frac{2}{49.51}}\)

Q=\(50.\frac{3\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)}{50\left(\frac{2}{1.99}+\frac{2}{3.97}+...+\frac{2}{49.51}\right)}\)

Q=\(50.3.\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}}\)

Q=\(150.\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{99+1}{1.99}+\frac{97+3}{3.97}+...+\frac{51+49}{49.51}}\)

Q=150\(.\frac{\frac{1}{1}+\frac{1}{3}+...+\frac{1}{99}}{\left(\frac{1}{1}+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)}\)

Q=\(150.\frac{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}\)

Q=150.1

Q=150

      \(Q=\frac{4+\frac{3}{5}+...+\frac{3}{95}+\frac{3}{97}+\frac{3}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{95.5}+\frac{1}{97.3}+\frac{1}{99.1}}\)

=> \(Q=\frac{100\left(\frac{3}{1}+\frac{3}{3}+\frac{3}{5}+...+\frac{3}{95}+\frac{3}{97}+\frac{3}{99}\right)}{100\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{95.5}+\frac{1}{97.3}+\frac{1}{99.1}\right)}\)

=> \(Q=\frac{100.3\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{95}+\frac{1}{97}+\frac{1}{99}\right)}{\frac{1+99}{1.99}+\frac{3+97}{3.97}+\frac{5+95}{5.95}+...+\frac{95+5}{95.5}+\frac{97+3}{97.3}+\frac{99+1}{99.1}}\)

=> \(Q=\frac{300\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{95}+\frac{1}{97}+\frac{1}{99}\right)}{\left(\frac{1}{1}+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{95}+\frac{1}{5}\right)+\left(\frac{1}{97}+\frac{1}{3}\right)+\left(\frac{1}{99}+\frac{1}{1}\right)}\)

=> \(Q=\frac{300\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{95}+\frac{1}{97}+\frac{1}{99}\right)}{2\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{95}+\frac{1}{97}+\frac{1}{99}\right)}\)

=> \(Q=\frac{300}{2}=150\)

Ko phải p/x mà là p/s muốn tỏ ra nguy hiểm à bn tú linh

tham khảo câu hỏi tương tự

Ta có : 

+) \(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)

\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{1.99}+\frac{100}{3.97}+\frac{100}{5.95}+...+\frac{100}{49.51}\)

\(=100\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}\right)\)

+) \(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}\)

\(=2\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}\right)\) 

\(\Rightarrow A=\frac{100\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}\right)}\)

\(\Rightarrow A=\frac{100}{2}\) 

\(\Rightarrow A=50\)

Tính ở tử số: 

\(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}=50.2.\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)\)

\(=50.\left(\frac{1}{1.99}+\frac{1}{3.97}+.....+\frac{1}{49.51}+\frac{1}{51.49}+...+\frac{1}{99.1}\right)\)

Gọi tử số là C: mẫu số là B => \(A=\frac{C}{A}=50\)

Kiệt này , có bài nào hay ko lôi hết ra xem. Mình sắp đi thi HSG toán 8 . Cậu xem có bài nào mà dễ vào ko bảo mình để mà mình ôn.

mình còn nhiều bài lắm nhưng lại hết 5 câu trong 1 ngày rồi

Ai có vip đăng hộ mình: (sắp thi rồi)

Bài 18:

Tính:

\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)