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\(\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}=\frac{x-4}{2016}\)
\(\Leftrightarrow\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}-\frac{x-4}{2016}=0\)
\(\Leftrightarrow\frac{x-1}{2019}-1+\frac{x-2}{2018}-1-\frac{x-3}{2017}+1-\frac{x-4}{2016}+1=0\)
\(\Leftrightarrow\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)
\(\Leftrightarrow x-2020=0\Leftrightarrow x=2020\)
\(\frac{x-1}{2019}+\frac{x-2}{2018}-\frac{x-3}{2017}=\frac{x-4}{2016}\)
\(\frac{x-1}{2019}+\frac{x-2}{2018}=\frac{x-3}{2017}+\frac{x-4}{2016}\)
\(\frac{x-1}{2019}+\frac{x-2}{2018}-2=\frac{x-3}{2017}+\frac{x-4}{2016}-2\)
\(\left(\frac{x-1}{2019}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-3}{2017}-1\right)+\left(\frac{x-4}{2016}-1\right)\)
\(\frac{x-1-2019}{2019}+\frac{x-2-2018}{2018}=\frac{x-3-2017}{2017}+\frac{x-4-2016}{2016}\)
\(\frac{x-2020}{2019}+\frac{x-2020}{2018}=\frac{x-2020}{2017}+\frac{x-2020}{2016}\)
\(\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)
\(\left(x-2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\right)=0\)
\(\Rightarrow x-2020=0\)
Vậy \(x=2020\)
Bài giải
a, \(\frac{x+5}{2017}-\frac{x+5}{2018}+\frac{x+5}{2019}-\frac{x+5}{2020}=0\)
\(\left(x+5\right)\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)=0\)
Do \(\left(\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}\right)\ne0\)
\(\Rightarrow\text{ }x+5=0\)
\(x=0-5\)
\(=-5\)
x+4/2015 + x+3/2016 = x+2/2017 + x+1/2018
=> 1 + x+4/2015 + 1 + x+3/2016 = 1 + x+2/2017 + 1 + x+1/2018
=> x+2019/2015 + x+2019/2016 = x+2019/2017 + x+2019/2018
=> x+2019/2015 + x+2019/2016 - x+2019/2017 - x+2019/2018 = 0
=> (x + 2019).(1/2015 + 1/2016 - 1/2017 - 1/2018) = 0
Vì 1/2015 > 1/2017; 1/2016 > 1/2018
=> 1/2015 + 1/2016 - 1/2017 - 1/2018 khác 0
=> x + 2019 = 0
=> x = -2019
Xin lỗi bạn nha dòng cuối mik nhầm ...
\(\Rightarrow x=-2018\)
Vậy x = -2018
\(\frac{x+4}{2014}+\frac{x+3}{2015}+\frac{x+2}{2016}+\frac{x+1}{2017}=a\)
\(\Rightarrow\frac{x+4}{2014}+1+\frac{x+3}{2015}+1+\frac{x+2}{2016}+1+\frac{x+1}{2017}+1=a+4\)
\(\frac{x+2018}{2014}+\frac{x+2018}{2015}+\frac{x+2018}{2016}+\frac{x+2018}{2017}=a+4\)
\(\left(x+2018\right).\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)=a+4\)
..
s bk
Câu a đề thiếu vế phải rồi bạn
b: \(\Leftrightarrow x\cdot0+1=0\)
=>0x+1=0(vô lý)
3. Tìm x biết: |15-|4.x||=2019
\(\Rightarrow\orbr{\begin{cases}15-\left|4x\right|=2019\\15-\left|4x\right|=-2019\end{cases}\Rightarrow\orbr{\begin{cases}\left|4x\right|=-2004\\\left|4x\right|=2034\end{cases}}}\)
vì \(4x\ge0\)\(\Rightarrow\)|4x|=2043\(\Rightarrow4x=2034\Rightarrow x=508,5\)
KL: x=508,5
đề thiếu hả bn
Hình như bạn ghi thiếu đề rồi. Để tìm đc x trong đẳng thức này thì ta phải có kết quả của biểu thức trên chứ đề cộc lốc thế này ko giải đc đâu