Giải phương trình: \(\dfrac{1}{x^2-x+1}+\dfrac{2}{x^2-x+2}+\dfrac{3}{x^2-x+3}+....+\dfrac{2018}{x^2-x+2018}=2018\)
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CC
1
26 tháng 1 2019
\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+2017}{3}+\frac{x+2016}{4}\)
\(\Leftrightarrow\frac{x+1}{2019}+1+\frac{x+2}{2018}+1=\frac{x+2017}{3}+1+\frac{x+2016}{4}+1\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}-\frac{x+2020}{3}-\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)=0\)
Mà \(\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)\ne0\)
\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)
Vậy...
13 tháng 3 2022
\(\dfrac{1}{x+1}\)-\(\dfrac{5}{x-2}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\)\(\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}\)-\(\dfrac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\)x-2-5(x+1)=15
\(\Leftrightarrow\) x-2-5x-5=15
\(\Leftrightarrow\)x-5x=15+2+5
\(\Leftrightarrow\)-4x=22
\(\Leftrightarrow\)x=-\(\dfrac{11}{2}\)
vậy
MM
1
27 tháng 1 2019
\(\dfrac{x+1}{2019}+\dfrac{x+2}{2018}=\dfrac{x+2017}{3}+\dfrac{x+2016}{4}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2019}+1\right)+\left(\dfrac{x+2}{2018}+1\right)=\left(\dfrac{x+2017}{3}+1\right)+\left(\dfrac{x+2016}{4}+1\right)\)
\(\Leftrightarrow\dfrac{x+2020}{2019}+\dfrac{x+2020}{2018}-\dfrac{x+2020}{3}-\dfrac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{3}-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x+2020=0\) ( do \(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{3}-\dfrac{1}{4}\ne0\))
\(\Leftrightarrow x=-2020\)
Vậy phương trình có tập nghiệm S = \(\left\{-2020\right\}\)
Y
3
23 tháng 9 2021
\(\Leftrightarrow\left(\dfrac{x+1}{2019}+1\right)+\left(\dfrac{x+2}{2018}+1\right)=\left(\dfrac{x+3}{2017}+1\right)+\left(\dfrac{x+4}{2016}+1\right)\)
\(\Leftrightarrow\dfrac{x+2020}{2019}+\dfrac{x+2020}{2018}-\dfrac{x+2020}{2017}-\dfrac{x+2020}{2016}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x=-2020\)(do \(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\ne0\))
TB
1
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 1 2019
\(\dfrac{2-x}{2017}+1=\dfrac{x-1}{2018}-1+1-\dfrac{x}{2019}\)
\(\Leftrightarrow\dfrac{2019-x}{2017}=\dfrac{x-2019}{2018}+\dfrac{2019-x}{2019}\)
\(\Leftrightarrow\dfrac{2019-x}{2017}+\dfrac{2019-x}{2018}-\dfrac{2019-x}{2019}=0\)
\(\Leftrightarrow\left(2019-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}\right)=0\)
\(\Leftrightarrow2019-x=0\) (do \(\dfrac{1}{2017}>\dfrac{1}{2019}\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}>0\))
\(\Rightarrow x=2019\)
NN
13 tháng 8 2023
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)
\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)
\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)
\(=>x+1=0\)
\(=>x=-1\)
b,
\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)
\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)
\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)
\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)
Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)
\(=>x+2021=0\)
\(=>x=-2021\)
c,
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)
\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)
\(=>x+329=0\)
\(=>x=-329\)
\(\dfrac{1}{x^2-x+1}+\dfrac{2}{x^2-x+2}+\dfrac{3}{x^2-x+3}+...+\dfrac{2018}{x^2-x+2018}=2018\)
\(\Leftrightarrow\left(\dfrac{1}{x^2-x+1}-1\right)+\left(\dfrac{2}{x^2-x+2}-1\right)+\left(\dfrac{3}{x^2-x+3}-1\right)+...+\left(\dfrac{2018}{x^2-x+2018}-1\right)=0\)
\(\Leftrightarrow\dfrac{1-x^2+x-1}{x^2-x+1}+\dfrac{2-x^2+x-2}{x^2-x+2}+\dfrac{3-x^2+x-3}{x^2-x+3}+...+\dfrac{2018-x^2+x-2018}{x^2-x+2018}=0\)
\(\Leftrightarrow-\left(x^2-x\right)\left(\dfrac{1}{x^2-x+1}+\dfrac{1}{x^2-x+2}+\dfrac{1}{x^2-x+3}+...+\dfrac{1}{x^2-x+2018}\right)=0\)
Ta có: \(\dfrac{1}{x^2-x+1}+\dfrac{1}{x^2-x+2}+...+\dfrac{1}{x^2-x+2018}>0\)
\(\Leftrightarrow x^2-x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)