\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{8\times9}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

\(=\frac{1}{1}-\frac{1}{9}=\frac{8}{9}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}\)

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

\(=\frac{8}{9}\)

   \(\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+\dfrac{1}{8\times9}+\dfrac{1}{9\times10}\)

= \(\dfrac{1}{5}-\dfrac{1}{6}\times\dfrac{1}{6}-\dfrac{1}{7}\times\dfrac{1}{7}-\dfrac{1}{8}\times\dfrac{1}{8}-\dfrac{1}{9}\times\dfrac{1}{9}-\dfrac{1}{10}\)

= \(\dfrac{1}{5}-\dfrac{1}{10}\)  

= \(\dfrac{1}{10}\)

Bài tập cuối tuần 10

=1/6-1/7+1/7-1/8+1/8-1/9+...+1/10-1/11

=1/6-1/11=5/66

`1/[4xx5]+1/[5xx6]+1/[6xx7]+1/[7xx8]+1/[8xx9]+1/[9xx10]`

`=1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10`

`=1/4-1/10`

`=10/40-4/40=6/40=3/20`

\(A=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}....\frac{1}{9x10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{9}-\frac{1}{10}=\frac{9}{10}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)

⇔ 3(10x+ 3) = 36+ 4(6 + 8x )

⇔ 30x + 9 = 36 + 24 + 32x

⇔ 30x - 32x = 36 + 24 – 9

⇔ -2x = 51

⇔ x = -25,5

Vậy phương trình có nghiệm x = -25,5

Dặt \(A=\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{2019.2020}\)

\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=\frac{1}{6}-\frac{1}{2020}\)

\(A=\frac{1007}{6060}\)

hok tốt!!

\(9\times7+8\times9-15\times9.\\ =9\times\left(7+8-15\right)=9\times0=0.\)

\(\Rightarrow\left(9\times7+8\times9-15\times9\right):\left(1+3+5+7+...+17+19\right).\\ =0:\left(1+3+5+7+...+17+19\right).\\ =0.\)

\(=\left(15\cdot9-15\cdot9\right):\left(1+3+...+17+19\right)=0\)

=9×(7+8-15)=9×0=0

=0:(1+3+5+7+...+17+19)

=0