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Sửa đề: \(\frac{x-3}{2018}\rightarrow\frac{x-3}{2016}\)
\(\frac{x-1}{2018}+\frac{x-2}{2017}=\frac{x-3}{2016}+\frac{x-4}{2015}\)
\(\Leftrightarrow\frac{x-1}{2018}-1+\frac{x-2}{2017}-1=\frac{x-3}{2016}-1+\frac{x-4}{2015}-1\)
\(\Leftrightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}=\frac{x-2019}{2016}+\frac{x-2019}{2015}\)
\(\Leftrightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}-\frac{x-2019}{2016}-\frac{x-2019}{2015}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x-2019=0\) (Vì \(\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)\ne0\) )
\(\Leftrightarrow x=2019\)
Vậy \(S=\left\{2019\right\}\)
Này Thục Trinh, chỗ mà \(\frac{x-1}{2018}-1+\frac{x-2}{2017}-1=\frac{x-3}{2016}-1+\frac{x-4}{2015}-1\)
bạn phải có đóng mở ngoặc vào chứ
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2002}{2}\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy : \(x=-2020\)
Chúc bạn học tốt !!
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\\ \left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\\ \frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\\ \frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\\ \left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\\ \Rightarrow x+2020=0\\ \Rightarrow x=-2020\)
Vậy x = -2020
b) \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\\ \left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)=\left(\frac{x+2017}{7}-1\right)+\left(\frac{x+2018}{8}-1\right)\\ \frac{x+2010}{5}+\frac{x+2010}{6}=\frac{x+2010}{7}+\frac{x+2010}{8}\\ \frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\\ \left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)
Vậy x = -2010
\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=3\)
\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=0\)
\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\right)=0\)
\(\Leftrightarrow x+2020=0\)( vì \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}>0\) )
\(\Leftrightarrow x=-2020\)
Vậy ...
\(\Rightarrow\frac{x-1}{2018}-1+\frac{x-2}{2017}-1=\frac{x-3}{2016}-1+\frac{x-4}{2015}-1\)
\(\Rightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}=\frac{x-2019}{2016}+\frac{x-2019}{2015}\)
\(\Rightarrow\orbr{\begin{cases}x=2019\left(1\right)\\\frac{1}{2018}+\frac{1}{2017}=\frac{1}{2016}+\frac{1}{2015}\left(2\right)\end{cases}}\) mà \(\left(2\right)\)không thể xảy ra nên x=2019 là nghiệm của phương trình.
ĐK \(2018x\ge0\Rightarrow x\ge0\)
Khi đó \(x+\frac{1}{2018}\ge0;x+\frac{2}{2018}\ge0;...;x+\frac{2017}{2018}\ge0\)
Ta có \(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|=2018x\)(Vế trái có 2017 hạng tử)
<=> \(x+\frac{1}{2018}+x+\frac{2}{2018}+...+x+\frac{2017}{2018}=2018x\)
<=> \(\left(x+x+...x\right)+\left(\frac{1}{2018}+\frac{2}{2018}+...+\frac{2017}{2018}\right)=2018x\)
2017 hạng tử x 2017 số hạng
=> \(2017x+\frac{1+2+...+2017}{2018}=2018x\)
=> \(x=\frac{2017.\left(2017+1\right):2}{2018}\)
\(\Rightarrow x=\frac{2017}{2}=1008,5\)(tm)
Vậy x = 1008,5
Vì \(\left|x+\frac{1}{2018}\right|\ge0\forall x\)
\(\left|x+\frac{2}{2018}\right|\ge0\forall x\)
\(\left|x+\frac{3}{2018}\right|\ge0\forall x\)
.......................................
\(\left|x+\frac{2017}{2018}\right|\ge0\forall x\)
\(\Rightarrow\)\(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+\left|x+\frac{3}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|\ge0\forall x\)
mà \(\left|x+\frac{1}{2018}\right|+\left|x+\frac{2}{2018}\right|+\left|x+\frac{3}{2018}\right|+...+\left|x+\frac{2017}{2018}\right|=2018x\)
\(\Rightarrow\)\(2018x\ge0\forall x\)\(\Rightarrow\)\(x\ge0\)
\(\Rightarrow\)\(x+\frac{1}{2018}+x+\frac{2}{2018}+x+\frac{3}{2018}+...+x+\frac{2017}{2018}=2018x\)
\(\Leftrightarrow\)\(2017x+\frac{1}{2018}+\frac{2}{2018}+\frac{3}{2018}+...+\frac{2017}{2018}=2018x\)
\(\Leftrightarrow\)\(\frac{1+2+3+...+2017}{2018}=x\)
\(\Leftrightarrow\)\(x=\frac{\left[\left(2017+1\right).2017\right]:2}{2018}\)
\(\Leftrightarrow\)\(x=\frac{2035153}{2018}\)
\(\Leftrightarrow\)\(x=\frac{2017}{2}=1008,5\)
Vậy \(x=1008,5\)