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

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

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

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

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

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

kết quả là 891/100 nha

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

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

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

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

\(=\frac{891}{100}\)
 

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

Ta có:

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

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

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

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

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

     \(=9.\frac{99}{100}\)

     \(=\frac{9.99}{100}\)

      \(=\frac{891}{100}\)

A=91.2+92.3+93.4+...+998.99+999.100

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

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

\(A=9.\left(1+0+0+...+0-\frac{1}{100}\right)\)

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

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

\(A=\frac{891}{100}=8\frac{91}{100}\)

k cho mk nha

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

\(A=\frac{9.1}{1.2.1}+\frac{9.1}{2.3.1}+...+\frac{9.1}{98.99.1}+\frac{9.1}{99.100.1}\)

\(A=1\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

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

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

\(A=1.\frac{99}{100}\)

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

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

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

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

=9-9/100

=891/100

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

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

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

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

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

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

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

\(=9\left(1-\frac{1}{100}\right)=9.\frac{99}{100}=\frac{891}{100}\)

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

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

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

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

\(=9.\left(1-\frac{1}{100}\right)=9.\frac{99}{100}=\frac{891}{100}\)

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