Question

* Tính:

\(\frac{1}{3.4}\) - \(\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}-\frac{1}{8.9}-\frac{1}{9.10}\)

- Help me. Please.

Answer 1

\(\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}-\frac{1}{8.9}-\frac{1}{9.10}\)

\(=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)

\(=\frac{1}{12}-\left[1\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\right]\)

\(=\frac{1}{12}-\frac{3}{20}\)

\(=-\frac{1}{15}\)

 

Answer 2

cộng thì còn lm dc mà trừ thì bt.com

Answer 3

7/30

 

Answer 4

- Dễ vồn thế này mà Hạo ko bít làm

Answer 5

\(\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}-\frac{1}{8.9}-\frac{1}{9.10}\)

\(=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}\right)\)

\(=\frac{1}{12}-\left[1\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\right]\)

\(=\frac{1}{12}-1.\left(\frac{3}{20}\right)\)

\(=\frac{1}{12}.\frac{3}{20}\)

\(=\frac{1}{80}\)

Answer 6

- ý t làm lộn cho làm lại

Answer 7

\(\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}-\frac{1}{8.9}-\frac{1}{9.10}\)

\(=\frac{1}{3.4}-\left(\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}-\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{12}-\left\{1\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\right\}\)

\(=\frac{1}{12}-\frac{3}{20}\\ =\frac{-1}{15}\)