Cách tìm BCNN:

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

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

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

ĐẶT : A= \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)\(\)
 

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)

= \(1-\frac{1}{99}=\frac{98}{99}\)

Gọi tổng đó là S 

TA có : S = \(\frac{1}{1.2}+\frac{1}{2.3}+......\frac{1}{98.99}+\frac{1}{99.100}\)

S = \(\frac{1}{1.2}-\frac{1}{99.100}=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)

Vậy S = \(\frac{4949}{9900}\)