999/1000(hình như v)

Áp dụng công thức \(\dfrac{1}{k\left(k+1\right)}=\dfrac{1}{k}-\dfrac{1}{k+1}\), ta có:

\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{999}-\dfrac{1}{1000}\right)=1-\dfrac{1}{1000}=\dfrac{999}{1000}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)

\(A=1-\dfrac{1}{1000}\)

\(A=\dfrac{999}{1000}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

\(=1-\frac{1}{1000}+1\)

\(=\frac{1000}{1000}-\frac{1}{1000}+\frac{1000}{1000}\)

\(=\frac{1999}{1000}\)

Tham khảo nhé~

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)

= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

= \(1-\frac{1}{1000}+1\)

= \(\frac{1999}{1000}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}\right)+1\)

\(=\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{999}-\frac{1}{1000}\right)+1\)

\(=\left(1-\frac{1}{1000}\right)+1\)

\(=\frac{999}{1000}+1\)

\(=\frac{1999}{1000}\)

Cách làm :

Áp dụng công thức : \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\)

\(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{999.1000}\)

\(\Leftrightarrow C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{999}-\dfrac{1}{1000}\)

\(\Leftrightarrow C=1-\dfrac{1}{1000}\)

\(\Leftrightarrow C=\dfrac{999}{1000}\)

\(F=\dfrac{1}{1.3}+\dfrac{1}{3.5}+.........+\dfrac{1}{99.101}\)

\(\Leftrightarrow2F=\dfrac{2}{1.3}+\dfrac{2}{3.5}+............+\dfrac{2}{99.101}\)

\(\Leftrightarrow2F=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{99}-\dfrac{1}{101}\)

\(\Leftrightarrow2F=1-\dfrac{1}{101}\)

\(\Leftrightarrow2F=\dfrac{100}{101}\)

\(\Leftrightarrow F=\dfrac{50}{101}\)

Giải:

\(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\)

\(\Leftrightarrow C=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)

\(\Leftrightarrow C=\dfrac{1}{1}-\dfrac{1}{1000}\)

\(\Leftrightarrow C=\dfrac{999}{1000}\)

Sửa đề:

\(F=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{999.1001}\)

\(\Leftrightarrow F=\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{999}-\dfrac{1}{1001}\right)\)

\(\Leftrightarrow F=\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{1001}\right)\)

\(\Leftrightarrow F=\dfrac{1}{2}.\dfrac{1000}{1001}\)

\(\Leftrightarrow F=\dfrac{500}{1001}\)

Chúc bạn học tốt!

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\)

=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)(Áp dụng t.c\(\dfrac{1}{a\left(a+1\right)}=\dfrac{1}{a}-\dfrac{1}{a+1}\))

=\(\dfrac{1}{1}-\dfrac{1}{1000}=\dfrac{999}{1000}\)

Vậy...

\(F=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{997.999}\)

=>\(2F=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{997.999}\)

=>\(2F=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{997}-\dfrac{1}{999}\)(áp dụng tính chất \(\dfrac{2}{a\left(a+2\right)}=\dfrac{1}{a}-\dfrac{1}{a+2}\))

=>\(2F=\dfrac{1}{1}-\dfrac{1}{999}=\dfrac{998}{999}\)

=>\(F=\dfrac{499}{999}\)

Vậy...

a, 1/1.2+1/2.3+1/3.4+...+1/999.1000

=  1/1-1/2+1/2-1/3+1/3-1/4+....+1/999-1/1000

=   1/1-1/1000

=   999/1000

b, 1/2.4+1/4.6+1/6.8+1/8.10

=  1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10

=  1/2-1/10

=   4/10  =2/5

a.

$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{1000-999}{999.1000}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}$

$=1-\frac{1}{1000}=\frac{999}{1000}$

b.

$5B=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+....+\frac{5}{495.500}$

$=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+....+\frac{500-495}{495.500}$

$=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{495}-\frac{1}{500}$

$=1-\frac{1}{500}=\frac{499}{500}$

$\Rightarrow B=\frac{499}{500}: 5= \frac{499}{2500}$

c.

$2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{998.999.100}$

$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{1000-998}{998.999.1000}$

$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{998.999}-\frac{1}{999.1000}$

$=\frac{1}{1.2}-\frac{1}{999.1000}=\frac{499499}{999000}$

$\Rightarrow C=\frac{499499}{999000}:2=\frac{499499}{1998000}$

1² /1.2 . 2²/2.3 . 3²/3.4 ...  999²/999.1000

= 1.1/1.2 . 2.2/2.3 . 3.3/3.4........... 999.999/999.1000

= 1/2. 2/3 . 3.4.....999/1000

= 1/1000

thanks

333333000 nhé bạn

 S = 1.2 + 2.3 + 3.4 + ... + 999.1000 
<=> 3S = 1.2.3 + 2.3.3 + 3.4.3 + ... + 999.1000.3 
xét 3.n.(n + 1) 
= 3n.(n + 1) 
= n.(n + 1)(n + 2 - n + 1) 
= n.(n + 1)(n + 2) - n(n - 1)(n + 1) 
thay vào S được 
3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 999.1000.1001 - 998.999.1000 
=> S = 999.1000.1001 ÷ 3 = 333333000

Đặt A= 1/1.2 + 1/2.3 + 1/3.4+...+ 1/999.1000

=1-1/2+1/2-1/3+1/3-1/4+...+1/999-1/1000

=1-1/1000

=999/1000

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{999.1000}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{999}-\dfrac{1}{1000}\)

\(=1-\dfrac{1}{1000}\)

\(=\dfrac{999}{1000}\)