Question

1.Tính nhanh:

\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

Ai nhanh mk tích cho !!

Answer 1

#)Giải :

\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

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

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

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

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

          #~Will~be~Pens~#

Answer 2

\(\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)

\(=-\left[\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\right]\)

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

\(=-\left[1-\frac{1}{100}\right]=-\frac{99}{100}\)

Answer 3

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Answer 4

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Answer 5

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Answer 6

   \(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

=\(\frac{1}{100.99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

= \(\frac{1}{100.99}-\left(\frac{1}{2.1}+\frac{1}{3.2}+...\frac{1}{97.98}+\frac{1}{98.99}\right)\)

=\(\frac{1}{100.99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)

= \(\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)

= \(\frac{1}{9900}-\frac{98}{90}\)

=\(\frac{1}{9900}-\frac{9800}{9900}\)

= \(\frac{-9799}{9900}\)

- Study well -

Answer 7

1/100*99 - 1/99*98 - 1/98*97 - .... - 1/3*2 - 1/2*1

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

= 1/9900 - (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/98 - 1/99)

= 1/9900 - (1 - 1/99)

= 1/9900 - 98/99

= ...

Answer 8

\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{100.99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(=\frac{1}{100.99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\right)\)

\(=\frac{1}{100.99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\right)\)

\(=\frac{1}{9900}-\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{9900}-\frac{98}{99}\)

\(=\frac{1}{9900}-\frac{9800}{9900}\)

\(=\frac{-9799}{9900}\)