Ta xét:

\(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}=\dfrac{3}{1.2.3.4};\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}=\dfrac{3}{2.3.4.5};.....;\dfrac{1}{7.8.9}-\dfrac{1}{8.9.10}=\dfrac{3}{7.8.9.10}\)

Gọi biểu thức phải tính là A, ta có:

3A=\(\dfrac{1}{1.2.3}-\dfrac{1}{8.9.10}=\dfrac{714}{4320}\)

Vậy A=\(\dfrac{238}{1440}\)

bạn giai tớ bài này

bài này không 1\299 + 4\299+7\299 +...... + 298\299

a)\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{1}{n+1}.\left(\frac{1}{n}-\frac{1}{n+2}\right)\)=\(\frac{1}{2}.\frac{1}{n\left(n+1\right)}-\frac{1}{2}.\frac{1}{\left(n+1\right)\left(n+2\right)}\)= \(\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\)

=> a = \(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)\)+\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{4}\right)\)+....+\(\frac{1}{2}\left(\frac{1}{2018}-\frac{1}{2019}\right)-\frac{1}{2}\left(\frac{1}{2019}-\frac{1}{2020}\right)\)=\(\frac{1}{2}\left(1-\frac{1}{2}\right)-\frac{1}{2}\left(\frac{1}{2019}-\frac{1}{2020}\right)\)=\(\frac{1}{4}\left(1-\frac{1}{2019.1010}\right)\)=\(\frac{2019.1010-1}{2.2019.2020}\)

b) tương tự \(\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}=\left(\frac{1}{n}-\frac{1}{n+1}\right)\left(\frac{1}{n+2}-\frac{1}{n+3}\right)\)=\(\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)-\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\)-\(\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)+\frac{1}{2}\left(\frac{1}{n+1}-\frac{1}{n+3}\right)\)=\(\frac{1}{6}\left(\frac{1}{n}-\frac{1}{n+1}\right)-\frac{1}{3}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\)+\(\frac{1}{6}\left(\frac{1}{n+2}-\frac{1}{n+3}\right)\)= M-P+N

Với n từ 1 đến 2017 thì

M= \(\frac{1}{6}\left(\frac{1}{1}-\frac{1}{2}\right)+\frac{1}{6}\left(\frac{1}{2}-\frac{1}{3}\right)+...\)+\(\frac{1}{6}\left(\frac{1}{2017}-\frac{1}{2018}\right)\)=\(\frac{1}{6}\left(1-\frac{1}{2018}\right)=\frac{2017}{6.2018}\)

N= \(\frac{1}{6}\left(\frac{1}{3}-\frac{1}{4}\right)+\frac{1}{6}\left(\frac{1}{4}-\frac{1}{5}\right)+...+\)\(\frac{1}{6}\left(\frac{1}{2019}-\frac{1}{2020}\right)=\)\(\frac{1}{6}\left(\frac{1}{3}-\frac{1}{2020}\right)=\frac{2017}{6.3.2020}\)

P= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3}\right)+\frac{1}{3}\left(\frac{1}{3}-\frac{1}{4}\right)+...+\)\(\frac{1}{3}\left(\frac{1}{2018}-\frac{1}{2019}\right)\)= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{2019}\right)=\frac{2017}{3.2.2019}\)

M+N-P = \(\frac{2017}{6}\left(\frac{1}{2018}+\frac{1}{3.2020}-\frac{1}{2019}\right)\)=\(\frac{2017}{6}.\left(\frac{1}{2018.2019}+\frac{1}{3.2020}\right)\)

=  \(\frac{2017\left(1010+1009.673\right)}{3.2018.2019.2020}\)

120 dm² x 5 + 4m² = 600 dm² +  4 m² = 6 m² + 4m² = 10 m²

=1/1.2-1/3.4+1/2.3-1/3.4+...+1/116.117-1/118.119

=1-1/2-1/3+1/4+1/2-1/3-1/3-1/4+...+1/116-1/117-1/118+1/119

=1+1/119=120/119(ko nhầm thì z)

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 45 + 10 = 55

* Hok tốt !

=(1+9)+(2+8)+(3+7)+(4+6)+10+5

=10+10+10+10+10+5

=10.5+5

=50+5

=55

1+2+3+4+5+6+7+8+9+10=(1+10)+(2+9)+(3+8)+(4+7)+(5+6)=11+11+11+11+11=11.5=55

`-2/4 = 1/(x-1)`

`=> -2(x-1)=4.1`

`=> -2(x-1)=4`

`=> x-1=4:(-2)`

`=> x-1=-2`

`=>x=-2+1`

`=>x=-1`