NT
1
K
Khách
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NT
0
HN
19 tháng 3 2021
\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}+\dfrac{x+3}{2018}+\dfrac{x+4}{2017}+4=0\)
⇔ \(\dfrac{x+1}{2020}+1+\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1+\dfrac{x+4}{2017}+1=0\)
\(\Leftrightarrow\) \(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}=0\)
⇔ \(\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\right)=0\)
\(Do\) \(\left(\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\right)\ne0\)
⇒ \(x+2021=0\)
⇔ \(x=-2021\)
\(Vậy\) \(x=-2021\)
TB
1
LA
11 tháng 3 2019
\(\Leftrightarrow\frac{x-3}{2016}-1=\left(\frac{x-2}{2017}-1\right)+\left(\frac{x-1}{2018}-1\right)\)
\(\Leftrightarrow\frac{x-3-2016}{2016}=\frac{x-2-2017}{2017}+\frac{x-1-2018}{2018}\)
\(\Leftrightarrow\frac{x-2019}{2016}-\frac{x-2019}{2017}-\frac{x-2019}{2018}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
Vì \(\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\) ( không tin cứ bấm máy tính mà xem =)) )
\(\Rightarrow x-2019=0\Rightarrow x=2019\)
TN
26 tháng 2 2018
\(\frac{x+1}{2018}-\frac{x+2}{2017}=\frac{x+3}{2016}+1\)
\(\Leftrightarrow\frac{x+1}{2018}+1-\left(\frac{x+2}{2017}+1\right)=\frac{x+3}{2016}+1\)
\(\Leftrightarrow\frac{x+2019}{2018}-\frac{x+2019}{2017}=\frac{x+2019}{2016}\)
\(\Leftrightarrow\left(x+2019\right)\left(\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)
Có: \(\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\ne0\)
\(\Leftrightarrow x+2019=0\Leftrightarrow x=-2019\)
Vậy...
L
2
19 tháng 12 2017
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+...+\dfrac{1}{\left(x+2017\right)\left(x+2018\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+...+\dfrac{1}{x+2017}-\dfrac{1}{x+2018}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+2018}\)
\(=\dfrac{2018}{x\left(x+2018\right)}\)
Dạng này mình làm suốt rồi, bạn cứ yên tâm.
Tính tổng
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2017)(x+2018)
Giải:\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+....+\frac{1}{\left(x+2017\right)\left(x+2018\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+..........+\frac{1}{x+2017}-\frac{1}{x+2018}\)
\(=\frac{1}{x}-\frac{1}{x+2018}\)
Vậy........................................