\(D=\dfrac{5}{1\cdot2}+...+\dfrac{5}{199\cdot200}\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\right)\)

\(=\dfrac{5}{2}\cdot\dfrac{199}{200}=\dfrac{199}{80}\)

Lời giải:

\(D=5\times \left(\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...+\frac{1}{199\times 200}\right)\)

\(=5\times \left(\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+...+\frac{200-199}{199\times 200}\right)\)

\(=5\times \left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{199}-\frac{1}{200}\right)=5\times (1-\frac{1}{200})\)

\(=5\times \frac{199}{200}=\frac{995}{200}=\frac{199}{40}\)

\(B=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{199\times200}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\)

\(=1-\dfrac{1}{200}=\dfrac{199}{200}\)

E=0,5 x 199/200=199/400

\(E=\dfrac{0.5}{1.2}+\dfrac{0.5}{2\cdot3}+...+\dfrac{0.5}{199\cdot200}\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{200}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{199}{200}=\dfrac{199}{400}\)

=\(\left(\dfrac{1}{4}-\dfrac{\left(\dfrac{13}{36}\right)}{\dfrac{1}{9}}:\left(\dfrac{2}{3}+\dfrac{\dfrac{7}{15}}{\dfrac{7}{30}}\right)\right)\)

= \(\left(\dfrac{1}{4}-\dfrac{13}{4}:\left(\dfrac{2}{3}+2\right)\right)\)

= \(\dfrac{1}{4}-\dfrac{13}{4}:\dfrac{8}{3}\)

= \(\dfrac{1}{4}-\dfrac{39}{32}\)

= \(-\dfrac{31}{32}\)

????? đề bài j vậy

`60% - 2 1/2 - 1/4 - 0,25:5/12`

`=3/5-5/2-1/4-1/4 . 12/5`

`=3/5-5/2-1/4-3/5`

`=(3/5-3/5)-(5/2+1/4)`

`=0-(10/4+1/4)`

`=-11/4`

`x+0,25=18/5+43/4`

`x+0,25=3,6+10,75`

`x=3,6+10,5`

`x=14,1`

Vậy `x=14,1`

   `x+0,25=18/5+43/4`

`⇔x+1/4=18/5+43/4`

`⇔ x-18/5=43/4-1/4`

`⇔ x-18/5=42/4=21/2`

`⇔ x-36/10=105/10`

`⇔ x=141/10`

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=1\\x-\dfrac{1}{3}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

\(\left|x-\dfrac{1}{3}\right|-0,25=\dfrac{3}{4}\)

\(\left|x-\dfrac{1}{3}\right|-\dfrac{1}{4}=\dfrac{3}{4}\)

\(\left|x-\dfrac{1}{3}\right|=\dfrac{3}{4}+\dfrac{1}{4}=1\)

\(\text{Vậy }x-\dfrac{1}{3}=1\)

       \(x\)         \(=1+\dfrac{1}{3}=\dfrac{4}{3}\)

\(\text{hoặc }x-\dfrac{1}{3}=-1\)

       \(x\)          \(=\left(-1\right)+\dfrac{1}{3}=\dfrac{-2}{3}\)

\(\Rightarrow x\in\left\{\dfrac{4}{3};\left(\dfrac{-2}{3}\right)\right\}\)

Ta có: \(\left(\dfrac{3}{5}+0.415-\dfrac{3}{200}\right)\cdot2\dfrac{2}{3}\cdot0.25\)

\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right)\cdot\dfrac{8}{3}\cdot\dfrac{1}{4}\)

\(=\dfrac{8}{3}\cdot\dfrac{1}{4}=\dfrac{8}{12}=\dfrac{2}{3}\)