\(\dfrac{x-2}{2016}+\dfrac{x-4}{1009}+\dfrac{x-6}{2020}=-4\)

<=>\(\dfrac{x-2}{2016}+1+\dfrac{x-4}{1009}+2+\dfrac{x-6}{2020}+1=0\)

<=>\(\dfrac{x+2014}{2016}+\dfrac{x+2014}{1009}+\dfrac{x+2014}{2020}=0\)

<=>\(\left(x+2014\right)\left(\dfrac{1}{2016}+\dfrac{1}{1009}+\dfrac{1}{2020}\right)=0\)

vì 1/2016+1/1009+1/2020 khác 0

=>x+2014=0<=>x=-2014

\(\Leftrightarrow\left(\dfrac{x+1}{2022}+1\right)+\left(\dfrac{x+3}{2020}+1\right)+\left(\dfrac{x+5}{2018}+1\right)+\left(\dfrac{x+7}{2016}+1\right)=0\)

=>x+2023=0

=>x=-2023

\(\dfrac{1}{x+1}\)-\(\dfrac{5}{x-2}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\)\(\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}\)-\(\dfrac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\)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

nhớ like nha

ta có : 

\(\frac{x-1009}{1001}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)

hay \(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\Leftrightarrow x-2010=0\)

hay x =2010

Vậy phương trình có nghiệm x = 2010

\(\frac{x-27}{1991}+\frac{x-60}{1958}+\frac{x}{1009}=4\)

<=> \(\frac{x-27}{1991}-1+\frac{x-60}{1958}-1+\frac{x}{1009}-2=0\)

<=> \(\frac{x-2018}{1991}+\frac{x-2018}{1958}+\frac{x-2018}{1009}=0\)

<=> x - 2018 = 0 

<=> x = 2018

Vậy:...

Lời giải:

a.

PT $\Leftrightarrow (x+3)^2=2016^{2020}-17^{91}+9$

Ta thấy: $2016^{2020}-17^{91}+9\equiv 0-(-1)^{91}+0\equiv -1\equiv 2\pmod 3$

Mà 1 scp thì chia $3$ chỉ dư $0$ hoặc $1$ nên pt vô nghiệm.

b.

$x^2=2016(y-1)^2-2017^{2019}\equiv 0-1^{2019}\equiv 3\pmod 4$
Mà 1 scp chia $4$ chỉ dư $0$ hoặc $1$ nên vô lý.

Vậy pt vô nghiệm.

c.

$(x-1)^2=2017^{2017}+1\equiv 1^{2017}+1\equiv 2\pmod 4$
Mà 1 scp khi chia cho $4$ chỉ dư $0$ hoặc $1$ nên vô lý

Vậy pt vô nghiệm

d.

$(x+2)^2=2018^{10}+4\equiv (-1)^{10}+1\equiv 2\pmod 3$

Mà 1 scp khi chia $3$ dư $0$ hoặc $1$ nên vô lý

Vậy pt vô nghiệm.

\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)

<=> \(\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)

<=> \(\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)

<=> \(\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)

<=> x + 2015 = 0  ( vì \(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\)) 

<=> x = - 2015 

Vậy x = -2015.

Giải phương trình :

\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)

\(\Rightarrow\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)

\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)

\(\Rightarrow\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)

Mà \(\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)>0\)

\(\Rightarrow x+2015=0\)

\(\Rightarrow x=-2015\)

\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)

\(\Rightarrow\left(\frac{x-2}{2017}+1\right)+\left(\frac{x-3}{2018}+1\right)=\left(\frac{x-4}{2019}+1\right)+\left(\frac{x-5}{2020}+1\right)\)

\(\Rightarrow\frac{x-2+2017}{2017}+\frac{x-3+2018}{2018}=\frac{x-4+2019}{2019}+\frac{x-5+2020}{2020}\)

\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}=\frac{x+2015}{2019}+\frac{x+2015}{2020}\)

\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)

\(\Rightarrow\left(x+15\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)

\(\Rightarrow x+2015=0\)

\(\Rightarrow x=-2015\)

Vậy x = - 2015