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

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

\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)

\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)

\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

\(A=\frac{B}{6}=\frac{100}{2}=50\)

Vậy \(A=50\)

6 ở đâu hả https://olm.vn/thanhvien/aihaibara0

a) Đặt B = \(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)

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

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

\(=100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right)\)

Đặt C = \(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\)

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

\(=2\cdot\frac{1}{1.99}+2\cdot\frac{1}{3.97}+...+2\cdot\frac{1}{49.51}\)

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

Thay B và C vào A 

\(\Rightarrow A=\frac{100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}=\frac{100}{2}=50\)

b) Đặt E = \(\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}\)

\(=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)

\(=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}\)

\(=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)

Thay E vào B

\(\Rightarrow B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}=\frac{1}{100}\)

a)50

b)1/100

tk ủng hộ nha

a,

\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}}\)

\(A=\frac{\left[1+\frac{1}{99}\right]+\left[\frac{1}{3}+\frac{1}{97}\right]+...+\left[\frac{1}{49}+\frac{1}{51}\right]}{2\left[\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right]}\)

\(A=\frac{\frac{100}{1.99}+\frac{100}{3.97}+\frac{100}{5.95}+...+\frac{100}{99.1}}{2\left[\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right]}\)

\(A=\frac{100\left[\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right]}{2\left[\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right]}=\frac{100}{2}=50\)

b, Ta có:

\(\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}=\left[1+\frac{98}{2}\right]+\left[1+\frac{97}{3}\right]+...+\left[1+\frac{1}{99}\right]+1\)

\(=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}=100\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right]\)

Thế vào:

\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right]}=\frac{1}{100}\)

Tử số=(1+1/3+1/5+1/7+...+1/97+1/99)x1/5 ={ ( 1+1/99) + ( 1/3 + 1/97 ) + ( 1/5 + 1/95) +.....+(1/49 + 1/51)} X 1/5 = (100/ 1 x 99 + 100/ 3 x 97 + 100/ 5 x 95 + ...+ 100/ 49 x 51)X 1/5 = ( 1/1x 99 + 1/ 3 x 97 + 1/ 5 x 95 +...+ 1/ 49 x 51) x 20 Mẫu số=2/1x99+2/3x97+2/5x95+...+2/49x51 = ( 1/1x 99 + 1/ 3 x 97 + 1/ 5 x 95 +...+ 1/ 49 x 51) x 2 Vậy phân số có giá trị = 20/2 = 10

Tử số=(1+1/3+1/5+1/7+...+1/97+1/99)x1/5
={ ( 1+1/99) + ( 1/3 + 1/97 ) + ( 1/5 + 1/95) +.....+(1/49 + 1/51)} X 1/5
= (100/ 1 x 99 + 100/ 3 x 97 + 100/ 5 x 95 + ...+ 100/ 49 x 51)X 1/5
= ( 1/1x 99 + 1/ 3 x 97 + 1/ 5 x 95 +...+ 1/ 49 x 51) x 20
Mẫu số=2/1x99+2/3x97+2/5x95+...+2/49x51
= ( 1/1x 99 + 1/ 3 x 97 + 1/ 5 x 95 +...+ 1/ 49 x 51) x 2
Vậy phân số có giá trị = 20/2 = 10