\(\dfrac{1}{2001.2003}+\dfrac{1}{2003.2005}+...+\dfrac{1}{2011.2013}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{2001.2003}+\dfrac{1}{2003.2005}+...+\dfrac{1}{2011.2013}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2001}-\dfrac{1}{2003}+\dfrac{1}{2003}-\dfrac{1}{2005}+...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2001}-\dfrac{1}{2013}\right)=\dfrac{2}{1342671}\)

=1/2.(2/2001.2003+2/2003.2005+.....+2/2011.2013)

=1/2(1/2001-1/2003+1/2003-1/2005+....+1/2011-1/2013)

=1/2(1/2001-1/2013)

=2/1342671

\(\dfrac{1.4+2.6+3.8+4.10+5.12}{5.2+10.3+15.4+20.5+25.6}\)

\(=\dfrac{2\left(1.2+2.3+3.4+4.5+5.6\right)}{5\left(1.2+2.3+3.4+4.5+5.6\right)}\)

\(=\dfrac{2}{5}\)

`#3107.101107`

\(y\times5-y\div5=15\\ y\times5-y\times\dfrac{1}{5}=15\\ y\times\left(5-\dfrac{1}{5}\right)=15\\ y\times\dfrac{24}{5}=15\\ y=15\div\dfrac{24}{5}\\ y=\dfrac{25}{8}\)

Vậy, `y=`\(\dfrac{25}{8}.\)

ko cs gia tri

Ta có:

5/6+7,7+5,9=[Số t1+ Số t2+ Số t3]*2

14,43333333=[số t1+số t2+số t3]

Suy ra:Số t1+ số t2+ số t3=14,43333333:2

Số t1+số t2+số t3=7,216666667

Đến đây dễ rồi, tự làm tiếp

Mình chỉ giúp thế thôi

3:(19/4 - x) = 4/5

=> 19/4 - x = 3: 4/5 = 15/4

=> x = 19/4 - 15/4 = 1

\(3:\left(\dfrac{19}{4}-x\right)=\dfrac{4}{5}\\ \dfrac{19}{4}-x=3:\dfrac{4}{5}\\ \dfrac{19}{4}-x=3\times\dfrac{5}{4}\\ \dfrac{19}{4}-x=\dfrac{15}{4}\\ x=\dfrac{19}{4}-\dfrac{15}{4}\\ x=\dfrac{4}{4}=1\)

\(840:100\times5\\ =840:500\\ =42\)

840:100*5

8,4*5

42

3/5 x = 5/6

x = 5/6 : 3/5

x = 25/18

\(\dfrac{3}{5}:x=\dfrac{1}{3}+\dfrac{1}{2}\)

\(\dfrac{3}{5}:x=\dfrac{5}{6}\)

\(x=\dfrac{3}{5}:\dfrac{5}{6}\)

\(x=\dfrac{18}{25}\)