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=> (x+2020)/5=(x+2020)/6=(x+2020)/3+(x+2020)/2
=>(x+2020)(1/5+1/6)=(x+2020)(1/3+1/2)
Với x+2020=0=>x=-2020
Với x+2020 khác 0=>1/5+1/6=1/3+1/2 ,vô lí
Vậy x=-2020
\(\frac{x-2017}{5}-\frac{x-2017}{6}=\frac{x-2017}{7}-\frac{x-2017}{8}\)
\(\frac{x-2017}{5}-\frac{x-2017}{6}-\frac{x-2017}{7}+\frac{x-2017}{8}=0\)
\(\left(x-2017\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\right)=0\)
mà \(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\ne0\)
\(\Rightarrow x-2017=0\)
\(\Rightarrow x=2017\)
Vậy x = 2017
\(\Rightarrow\frac{x-2017}{5}-\frac{x-2017}{6}-\frac{x-2017}{7}+\frac{x-2017}{8}=0\)
\(\Rightarrow\left(x-2017\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\right)=0\)
\(\Rightarrow x-2017=0\)(vì \(\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}\ne0\))
=>x=2017
Chúc bạn học tốt
\(\dfrac{x-2017}{2019}+\dfrac{x-2019}{2017}=\dfrac{x+6}{2021}\)
\(\Rightarrow\dfrac{x-2017}{2019}-1+\dfrac{x-2019}{2017}-1=\dfrac{x+6}{2021}-2\)
\(\Rightarrow\dfrac{x-2017}{2019}-\dfrac{2019}{2019}+\dfrac{x-2019}{2017}-\dfrac{2017}{2017}=\dfrac{x+6}{2021}-\dfrac{4042}{2021}\)
\(\Rightarrow\dfrac{x-2017-2019}{2019}+\dfrac{x-2019-2017}{2017}=\dfrac{x+6-4042}{2021}\)
\(\Rightarrow\dfrac{x-4036}{2019}+\dfrac{x-4036}{2017}=\dfrac{x-4036}{2021}\)
\(\Rightarrow\dfrac{x-4036}{2021}-\dfrac{x-4036}{2019}-\dfrac{x-4036}{2017}=0\)
\(\Rightarrow\left(x-4036\right)\left(\dfrac{1}{2021}-\dfrac{1}{2019}-\dfrac{1}{2017}\right)=0\)
=> x - 4036 = 0
=> x = 4036
x − 2017/2019 + x−2019/2017 = x+6/2021
=> x − 2017/2019 + x−2019/2017 = x+6/2021
=> x − 2017/2019 − 1 + x − 2019/2017 − 1 = x + 6/2021 − 2
=> x − 2017/2019 − 1 + x − 2019/2017 − 1 = x + 6/2021 − 2
=> x − 2017/2019 − 2019/2019 + x − 2019/2017 − 2017/2017
= x + 6/2021 − 4042/2021
=> x − 2017/2019 − 2019/2019 + x − 2019/2017 − 2017/2017
= x + 6/2021 − 4042/2021
=> x − 2017 − 2019/ 2019 + x − 2019 − 2017/2017
= x + 6 − 4042/2021
=> x − 2017 − 2019/2019 + x − 2019 − 2017/2017 = x + 6 − 4042/2021
=> x − 4036/2019 + x − 4036/2017 = x − 4036/2021
=> x − 4036/2019 + x − 4036/2017 = x − 4036/2021
=> x − 4036/2021 − x − 4036/2019 − x − 4036/2017 = 0
=> x − 4036/2021 − x − 4036/2019 − x − 4036/2017 = 0
=>(x − 4036)(12021 − 12019 − 12017) = 0
=> x - 4036 = 0
=> x = 4036
Lâp bảng xét dấu
2016 2017
x-2016 _ 0 + +
x-2017 _ _ 0 +
Nếu x<2016 thì |x-2016|=2016-x,|x-2017|=2017-x
Ta có 2016-x+2017-x=2018
4033-2x=2018
2x=2015
x=1007,5
Nếu 2016<=x<=2017thif |x-2016|=x-2016;|x-2017|=2017-x
Ta có x-2016+2017-x=2018
ox+1=2018
0x=2017 (vô lí)
Nếu x>=2017 thi |x-2016|=x-2016;|x-2017|=x-2017
Ta có x-2016+x-2017=2018
2x-4033=2018
2x=6051
x=3025,5
Vậy x=1007,5 hoăc x=3025,5
\(x\left(6-x\right)^{2017}=\left(6-x\right)^{2017}\)
\(x\left(6-x\right)^{2017}-\left(6-x\right)^{2017}=0\)
\(\left(x-1\right)\left(6-x\right)^{2017}=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(6-x\right)^{2017}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\6-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=6\end{cases}}\)
vậy \(\orbr{\begin{cases}x=1\\x=6\end{cases}}\)