\(\dfrac{x-1}{2023}+\dfrac{x-2}{2022}=\dfrac{x-3}{2021}+\dfrac{x-4}{2020}\)

`<=>(x-1)/2023-1+(x-2)/2022-1=(x-3)/2021-1+(x-4)/2020-1`

`<=>(x-2024)/2023+(x-2024)/2022=(x-2024)/2021+(x-2024)/2020`

`<=>(x-2024)(1/2023+1/2022-1/2021-1/2020)=0`

`<=>x-2024=0(1/2023+1/2022-1/2021-1/2020>0)`

`<=>x=2024`

=>\(\left(\dfrac{x-1}{2023}-1\right)+\left(\dfrac{x-2}{2022}-1\right)=\left(\dfrac{x-3}{2021}-1\right)+\left(\dfrac{x-4}{2020}-1\right)\)

=>x-2024=0

=>x=2024

\(\dfrac{x-1}{2023}+\dfrac{x-2}{2022}=\dfrac{x-3}{2021}+\dfrac{x-4}{2020}\)

⇔\(\dfrac{x-1}{2023}-1+\dfrac{x-2}{2022}-1=\dfrac{x-3}{2021}-1+\dfrac{x-4}{2020}\)

⇔\(\dfrac{x-1}{2023}-\dfrac{2023}{2023}+\dfrac{x-2}{2022}-\dfrac{2022}{2022}=\dfrac{x-3}{2021}-\dfrac{2021}{2021}+\dfrac{x-4}{2020}-\dfrac{2020}{2020}\)

⇔\(\dfrac{x-2024}{2023}+\dfrac{x-2024}{2022}=\dfrac{x-2024}{2021}+\dfrac{x-2024}{2020}\)

⇔\(\dfrac{x-2024}{2023}+\dfrac{x-2024}{2022}-\dfrac{x-2024}{2021}-\dfrac{x-2024}{2020}=0\)

⇔\(\left(x-2024\right)\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\ne0\right)\)

⇔\(x-2024=0\)

⇔\(x=2024\)

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

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

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

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

<=> 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 

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

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

\(\Leftrightarrow\)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

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.

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

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

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

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

\(\Rightarrow x+2015=0\Rightarrow x=-2015\)

\(S=\left\{-2015\right\}\)

gợi ý 

2017-x-2=2018-3-x=2019-4-x=2020-5-x

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

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

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

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

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

\(\Leftrightarrow x+2015=0\)

\(\Leftrightarrow x=-2015\)

helpppppppppppppppp

Bài làm:

Pt <=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-3}{2018}-1\right)+\left(\frac{x-5}{2016}-1\right)+\left(\frac{x-7}{2014}-1\right)=4-4\)

\(\Leftrightarrow\frac{x-2021}{2020}+\frac{x-2021}{2018}+\frac{x-2021}{2016}+\frac{x-2021}{2014}=0\)

\(\Rightarrow x-2021=0\Rightarrow x=2021\)

Ta có :\(\frac{x-1}{2020}+\frac{x-3}{2018}+\frac{x-5}{2016}+\frac{x-7}{2014}=4\)

=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-3}{2018}-1\right)+\left(\frac{x-5}{2016}-1\right)+\left(\frac{x-7}{2014}-1\right)=0\)

=> \(\frac{x-2021}{2020}+\frac{x-2021}{2018}+\frac{x-2021}{2016}+\frac{x-2021}{2014}=0\)

=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2018}+\frac{1}{2016}+\frac{1}{2014}\right)=0\)

Vì \(\frac{1}{2020}+\frac{1}{2018}+\frac{1}{2016}+\frac{1}{2014}\ne0\)

=> x - 2021 = 0

=> x = 2021

Vậy x = 2021

\(\dfrac{x+1}{2020}+\dfrac{x-1}{2018}=\dfrac{x+5}{2024}+\dfrac{x-5}{2014}\)

\(\Leftrightarrow\left(\dfrac{x+1}{2020}-1\right)+\left(\dfrac{x-1}{2018}-1\right)-\left(\dfrac{x+5}{2024}-1\right)-\left(\dfrac{x-5}{2014}-1\right)=0\)

\(\Leftrightarrow\dfrac{x-2019}{2020}+\dfrac{x-2019}{2018}-\dfrac{x-2019}{2024}-\dfrac{x-2019}{2014}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\dfrac{1}{2020}+\dfrac{1}{2018}-\dfrac{1}{2024}-\dfrac{1}{2014}\right)=0\)

\(\Leftrightarrow x-2019=0\\ \Leftrightarrow x=2019\)

a, Làm

\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x+5}{2016}+\frac{x+6}{2015}\)

<=>\(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2016}+\frac{x+2021}{2015}\)

<=>\(\left(x+2021\right)\left(\frac{1}{2020}+\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)

<=> x+2021=0

<=> x=-2021

Kl:......................

b, Làmmmmm

\(\frac{2-x}{2004}-1=\frac{1-x}{2005}-\frac{x}{2006}\)

<=> \(\frac{2006-x}{2004}=\frac{2006-x}{2005}+\frac{2006-x}{2006}\)

<=> \(\left(2006-x\right)\left(\frac{1}{2004}-\frac{1}{2005}-\frac{1}{2006}\right)=0< =>2006-x=0\)

<=> x=2006

Kl:..............