A = 3/2x4/3x5/4x....x100/99=100/2=50

A=(1/2+1)*(1/3+1)*(1/4+1).....(1/99+1)

A=3/2*4/3*5/4.....100/99    (Thực hiện tính tổng trong mỗi ngoặc đơn)

A=(3*4*5...100)/(2*3*4...99)

A=100/2                          (Rút gọn những thừa số giống nhau ở tử và mẫu)

A=50

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Sửa:\(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right).....\left(\frac{1}{99}+1\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

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

cách viết phân số trên máy tính kiểu gì vậy nhỉ

A = ( 1/2 + 1 )( 1/3 + 1)...(1/99 + 1)

= ( 1/2 + 2/2)(1/3 + 3/3)... (1/99 + 99/99)

= 3/2 . 4/3 ... 100/99

= 3 . 4 ... 100 / 2 . 3 . 99

= 100/2 = 50

                     #Louis

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

\(A=\frac{3}{2}.\frac{4}{3}.............\frac{100}{99}=\frac{3.4....................100}{2.3.................99}=\frac{\left(3.4.......99\right).100}{2.\left(3.4...........99\right)}=\frac{100}{2}=50\)

Vậy A=50

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

=\(\frac{3}{2}.\frac{4}{3}.............\frac{100}{99}\)

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

=50 nha bn

Cách tìm BCNN:

(1 - 1/2)(1 - 1/3)(1 - 1/4) ... (1 - 1/99) 

= 1/2*2/3*3/4*...*98/99

= 1/99

Ta có : (1 -1/2)(1-1/3)(1-1/4)..(1-1/99)

=1/2 .2/3.3/4....98/99

=1/99

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{98}{99}\)

\(=\frac{1\cdot2\cdot3.....98}{2\cdot3\cdot4.....99}=\frac{1}{99}\)

\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}=\dfrac{100}{2}=50\)

a)

C = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = 1 − 2 + 3 − 4 + ... + 97 − 98 + 99 − 100 = − 1 + − 1 + ... + − 1 + − 1 = − 1.50 = − 50.

b)

B = 1 − 2 − 3 + 4 + 5 − 6 − 7 + ... + 97 − 98 − 99 + 100 = 1 − 2 + − 3 + 4 + 5 − 6 + ... + 97 − 98 + − 99 + 100 = − 1 + 1 + − 1 + ... + − 1 + 1 = − 1 + 1 + − 1 + 1 + ... + − 1 + 1 − 1 = 0 + 0 + ... + 0 − 1 = − 1.

Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

=100

Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)