H=\(\dfrac{0,25}{1x2}\)+\(\dfrac{0,25}{2x3}\)+......+\(\dfrac{0,25}{199x200}\)
K
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NT
2
6 tháng 10 2021
\(D=\dfrac{5}{1\cdot2}+...+\dfrac{5}{199\cdot200}\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{199}{200}=\dfrac{199}{80}\)
AH
Akai Haruma
Giáo viên
6 tháng 10 2021
Lời giải:
\(D=5\times \left(\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...+\frac{1}{199\times 200}\right)\)
\(=5\times \left(\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+...+\frac{200-199}{199\times 200}\right)\)
\(=5\times \left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{199}-\frac{1}{200}\right)=5\times (1-\frac{1}{200})\)
\(=5\times \frac{199}{200}=\frac{995}{200}=\frac{199}{40}\)
6 tháng 10 2021
\(E=\dfrac{0.5}{1.2}+\dfrac{0.5}{2\cdot3}+...+\dfrac{0.5}{199\cdot200}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{200}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{199}{200}=\dfrac{199}{400}\)
DH
20 tháng 11 2018
=\(\left(\dfrac{1}{4}-\dfrac{\left(\dfrac{13}{36}\right)}{\dfrac{1}{9}}:\left(\dfrac{2}{3}+\dfrac{\dfrac{7}{15}}{\dfrac{7}{30}}\right)\right)\)
= \(\left(\dfrac{1}{4}-\dfrac{13}{4}:\left(\dfrac{2}{3}+2\right)\right)\)
= \(\dfrac{1}{4}-\dfrac{13}{4}:\dfrac{8}{3}\)
= \(\dfrac{1}{4}-\dfrac{39}{32}\)
= \(-\dfrac{31}{32}\)
WN
1
CT
1
17 tháng 5 2021
`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`
17 tháng 5 2021
`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`
2 tháng 1 2022
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=1\\x-\dfrac{1}{3}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2 tháng 1 2022
\(\left|x-\dfrac{1}{3}\right|-0,25=\dfrac{3}{4}\)
\(\left|x-\dfrac{1}{3}\right|-\dfrac{1}{4}=\dfrac{3}{4}\)
\(\left|x-\dfrac{1}{3}\right|=\dfrac{3}{4}+\dfrac{1}{4}=1\)
\(\text{Vậy }x-\dfrac{1}{3}=1\)
\(x\) \(=1+\dfrac{1}{3}=\dfrac{4}{3}\)
\(\text{hoặc }x-\dfrac{1}{3}=-1\)
\(x\) \(=\left(-1\right)+\dfrac{1}{3}=\dfrac{-2}{3}\)
\(\Rightarrow x\in\left\{\dfrac{4}{3};\left(\dfrac{-2}{3}\right)\right\}\)
1
7 tháng 7 2021
Ta có: \(\left(\dfrac{3}{5}+0.415-\dfrac{3}{200}\right)\cdot2\dfrac{2}{3}\cdot0.25\)
\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right)\cdot\dfrac{8}{3}\cdot\dfrac{1}{4}\)
\(=\dfrac{8}{3}\cdot\dfrac{1}{4}=\dfrac{8}{12}=\dfrac{2}{3}\)
\(H=0,25\times\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{19\times20}\right)\)
\(=0,25\times\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(=0,25\times\left(1-\dfrac{1}{20}\right)=0,25\times\dfrac{19}{20}=\dfrac{19}{80}\)
\(H=\dfrac{0.25}{1\cdot2}+\dfrac{0.25}{2\cdot3}+...+\dfrac{0.25}{199\cdot200}\)
\(=\dfrac{1}{4}\cdot\dfrac{199}{200}=\dfrac{199}{800}\)