a = 9/1.2 + 9/2.3 + 9/3.4 + ... + 9/98.99 + 9/99.100

a = 9.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/98.99 + 1/99.100)

a = 9.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/98 - 1/99 + 1/99 - 1/100)

a = 9.(1 - 1/100)]

a = 9.99/100

a = 891/100

\(a=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
      \(=9.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
      \(=9.\left(1-\frac{1}{100}\right)\)
      \(=9.\)\(\frac{99}{100}\)
      \(=\frac{891}{100}\)

a = 9/1.2 + 9/2.3 + 9/3.4 + ... + 9/98.99 + 9/99.100

a = 9.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/98.99 + 1/99.100)

a = 9.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/98 - 1/99 + 1/99 - 1/100)

a = 9.(1 - 1/100)]

a = 9.99/100

a = 891/100

A=\(\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)

A=9(\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}+\frac{1}{99.100}\))

A=9(\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\))

A=9(\(1-\frac{1}{100}\))

=9.\(\frac{99}{100}\)

=\(\frac{891}{100}\)

bạn làm đúng

lollllolololl

\(A=\dfrac{9}{1.2}+\dfrac{9}{2.3}+...+\dfrac{9}{99.100}\)

\(=9\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)

\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=9\left(1-\dfrac{1}{100}\right)\)

\(=9.\dfrac{99}{100}\)

\(=\dfrac{891}{100}\)

Vậy \(A=\dfrac{891}{100}\)

\(A=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)

\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(=9\cdot\dfrac{98}{99}=\dfrac{98}{11}\)

A = 9/1.2 + 9/2.3 + 9/3.4 +...+ 9/98.99 + 9/99.100

   = 9. (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/98 - 1/99 + 1/99 - 1/100)

   = 9. (1 - 1/100)

   = 9 . 99/100

   = 891/100

\(A=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=9\left(1-\dfrac{1}{100}\right)=\dfrac{891}{100}\)

\(A=\dfrac{9}{1.2}+\dfrac{9}{2.3}+\dfrac{9}{3.4}+...+\dfrac{9}{98.99}+\dfrac{9}{99.100}\)
\(=9\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=9\left(1-\dfrac{1}{100}\right)\)
\(=9.\dfrac{99}{100}\)
\(=\dfrac{891}{100}\)

\(A=\dfrac{9}{1.2}+\dfrac{9}{2.3}+\dfrac{9}{3.4}+...+\dfrac{9}{98.99}+\dfrac{9}{99.100}\)

\(=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=9.\left(1-\dfrac{1}{100}\right)\)

\(=9.\dfrac{99}{100}\)

\(=\dfrac{891}{100}\).

\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)

\(A=9\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=9\cdot\left(1-\frac{1}{100}\right)\)

\(A=9\cdot\frac{99}{100}\)

\(A=\frac{891}{100}\)

ta có:

\(=9.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

\(=9.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=9.\left(1-\frac{1}{100}\right)\)

\(=9.\frac{99}{100}\)

\(=\frac{11}{100}\)

 mk xin lỗi, mk bị nhầm:

kq=\(\frac{891}{100}\)

\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

\(A=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=9\left(1-\frac{1}{100}\right)\)

\(A=9\times\frac{99}{100}\)

\(A=\frac{891}{100}\) hoặc =8,91

A=9/1.2+9/2.3+9/3.4+...+9/98.99+9/99.100

A=9.(1/1.2+1/2.3+1/3.4+...+1/98.99+1/99.100)

A=9.(1/1-1/2+1/2-1/3+1/3-1/4+...+1/98-1/99+1/99-1/100)

A=9.(1/1-1/100)

A=9.99/100

A=891/100

A=8+91/100 ( viết dưới dạng hỗn số )

Vậy A=8+91/100

Nkớ k cho mink đó nha !!!