\(A=4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\)

\(=4\cdot\dfrac{2014}{2015}=\dfrac{8056}{2015}\)

`A=4/(1.2)+4/(2.3)+4/(3.4)+......+4/(2014.2015)`
`=4(1/(1.2)+1/(2.3)+1/(3.4)+......+1/(2014.2015))`
`=4(1-1/2+1/2-1/3+1/3-1/4+....+1/2014-1/2015)`
`=4(1-1/2015)`
`=4. 2014/2015`
`=8056/2015`

A=4.(1/1.2+1/2.3+...+1/2014.2015)

A=4.(1-1/2+1/2-1/3+...+1/2014-1/2015)

A=4.(1-1/2015)

A=4.2014/2015

A=8056/2015

Giải:

\(A=\dfrac{4}{1.2}+\dfrac{4}{2.3}+\dfrac{4}{3.4}+...+\dfrac{4}{2014.2015}\) 

\(A=4.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2014.2015}\right)\) 

\(A=4.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\) 

\(A=4.\left(\dfrac{1}{1}-\dfrac{1}{2015}\right)\) 

\(A=4.\dfrac{2014}{2015}\) 

\(A=\dfrac{8056}{2015}\)

\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2014.2015}\)

\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)

\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(\Leftrightarrow\frac{1}{4}A=\frac{1}{1}-\frac{1}{2015}\)

\(\Leftrightarrow\frac{1}{4}A=\frac{2014}{2015}\)

\(\Leftrightarrow A=\frac{2014}{2015}\div\frac{1}{4}\)

\(\Leftrightarrow A=\frac{8056}{2015}\)

\(=4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)=4\cdot\dfrac{2014}{2015}=\dfrac{8056}{2015}\)

\(\dfrac{4}{1.2}+\dfrac{4}{2.3}+...+\dfrac{4}{2014.2015}\\ =4\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2014.2015}\right)\\ =4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\\ =4\left(1-\dfrac{1}{2015}\right)\\ =4.\dfrac{2014}{2015}\\ =\dfrac{8056}{2015}\)

\(\dfrac{8056}{2015}\)

Mình k ghi lại đề nhé!~~

\(A=\dfrac{4}{1}.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2014.2015}\right)\)

\(A=\dfrac{4}{1}.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\right)\)

\(A=\dfrac{4}{1}.\left(1-\dfrac{1}{2015}\right)\)

\(A=\dfrac{4}{1}.\left(\dfrac{2015-1}{2015}\right)\)

\(A=\dfrac{4}{1}.\dfrac{2014}{2015}\)

\(A=3,998014888\)

\(A\approx4\)

Ta có: \(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2014.2015}\)

\(=1\left(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2014.2015}\right)\)

\(=1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)

\(=1\left(1-\frac{1}{2015}\right)\)

\(=1\left(\frac{2015}{2015}-\frac{1}{2015}\right)-1\left(\frac{2014}{2015}\right)=\frac{2014}{2015}\)

Vậy.....

\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+....+\frac{4}{2014.2015}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2014}-\frac{1}{2015}\)

\(A=\frac{1}{1}-\frac{1}{2015}=\frac{2015}{2015}-\frac{1}{2015}=\frac{2014}{2015}\)

Ta có: 

A = \(\frac{4}{1.2}+\frac{4}{2.3}+...+\frac{4}{2014.2015}\)

=\(4.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}\right)\)

=\(4.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)

=\(4.\left(\frac{1}{1}-\frac{1}{2015}\right)\)

=\(4.\frac{2014}{2015}\)

=\(\frac{8056}{2015}\)

cau hỏi tương tự ko có mà!!!!!!!!!!!!!!!!!!!!!!!!!!!!

3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+2014.2015.(2016-2013)

3C=2014.2015.2016

C=2014.2015.2016:3

D=673015

E=2550

C=...

G=781875

lấy lời giải ra hộ  tui

D=1+4+7+...+2008
D=(2008-1):3+1
D=670.
VẬY D=670.
E=2+4+6+...+100
E=(100-2):2+1
E=49.
VẬY E=49.
C=2+4+6+...+1024
C=(1024-2):2+1
C=512.
VẬY C=512.
 

\(A=\frac{4}{1.2}+\frac{4}{2.3}+...+\frac{4}{2014.2015}\)

\(A=4\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}\right)\)

\(A=4\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)

\(A=4\left(\frac{1}{1}-\frac{1}{2015}\right)\)

\(A=4\left(\frac{2015-1}{2015}\right)\)

\(A=4.\frac{2014}{2015}\)

... BẠN TỰ LÀM NỐT NHÉ!