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

Lời giải:

** Sửa đề: Chỗ $\frac{1}{1}$ ở mẫu chuyển thành $\frac{1}{2}$

$\frac{1}{1}.99+\frac{1}{3}.97+\frac{1}{5}.95+....+\frac{1}{97}.3+\frac{1}{99}.1$

$=50+(\frac{97}{3}+1)+(\frac{95}{5}+1)+....+(\frac{3}{97}+1)+(\frac{1}{99}+1)$

$=50+\frac{100}{3}+\frac{100}{5}+...+\frac{100}{97}+\frac{100}{99}$
$=100(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})$

\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{100(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})}=\frac{1}{100}\)

\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.99}+...+\frac{1}{99.1}}\)

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

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

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

\(=\frac{100}{2}=50\)

50 nha

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