Tham khảo :

Nhận thấy vế trái luôn dương nên \(x-2020\ge0\Leftrightarrow x\ge2020\)

Với \(x\ge2020\Rightarrow\left\{{}\begin{matrix}x-2017\ge0\\2x-2018\ge0\\3x-2019\ge0\end{matrix}\right.\)

PT trở thành: \(x-2017+2x-2018+3x-2019=x-2020\)

Hay kết hợp với điều kiện \(x=\dfrac{4034}{5}\) suy ra PT đã cho vô nghiệm 

\(\left|x-2017\right|+\left|2x-2018\right|+\left|3x-2019\right|=x-2020\)

\(ĐK:x\ge2020\)

\(\Leftrightarrow x-2017+2x-2018+3x-2019=x-2020\)

\(\Leftrightarrow5x=4034\)

\(\Leftrightarrow x=806,8\left(tm\right)\)

Vậy \(S=\left\{806,8\right\}\)

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

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

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

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

<=> 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 

Đề bài yêu cầu gì?

\(1,\\ b,\Leftrightarrow\left(x^2+4x+4\right)+\left(y-1\right)^2=25\\ \Leftrightarrow\left(x+2\right)^2+\left(y-1\right)^2=25\)

Vậy pt vô nghiệm do 25 ko phải tổng 2 số chính phương

\(2,\\ a,\Leftrightarrow x^2-\left(y^2-6y+9\right)=47\\ \Leftrightarrow x^2-\left(y-3\right)^2=47\)

Mà 47 ko phải hiệu 2 số chính phương nên pt vô nghiệm

\(b,\Leftrightarrow\left(x-2\right)^2+\left(3y-1\right)^2=16\)

Mà 16 ko phải tổng 2 số chính phương nên pt vô nghiệm

1a. Đề lỗi

1b. 

PT $\Leftrightarrow (x+2)^2+(y-1)^2=25$

$\Leftrightarrow (x+2)^2=25-(y-1)^2\leq 25$

$(x+2)^2$ là scp không vượt quá $25$ nên có thể nhận các giá trị $0,1,4,9,16,25$

Nếu $(x+2)^2=0\Rightarrow (y-1)^2=25$

$\Rightarrow (x,y)=(-2, 6), (-2, -4)$
Nếu $(x+2)^2=1\Rightarrow (y-1)^2=24$ không là scp (loại)

Nếu $(x+2)^2=4\Rightarrow (y-1)^2=21$ không là scp (loại)

Nếu $(x+2)^2=9\Rightarrow (y-1)^2=16$

$\Rightarrow (x,y)=(1, 5), (1, -3), (-5,5), (-5, -3)$

Nếu $(x+2)^2=25\Rightarrow (y-1)^2=0$

$\Rightarrow (x,y)=(3, 1), (-7, 1)$

1c. 

Vì $x^2$ là scp nên $x^2\equiv 0,1\pmod 3$

$3(y-1)^2\equiv 0\pmod 3$

$\Rightarrow x^2+3(y-1)^2\equiv 0,1\pmod 3$

Mà $2021\equiv 2\pmod 3$
Do đó pt $x^2+3(y-1)^2=2021$ vô nghiệm

1d.

Ta thấy:

$(3x-1)^{2020}$ là scp không chia hết cho $3$ nên $(3x-1)^{2020}\equiv 1\pmod 3$

$18(y-2)^{2019}\equiv 0\pmod 3$

$\Rightarrow (3x-1)^{2020}+18(y-2)^{2019}\equiv 1\pmod 3$
Mà $2019^{2020}\equiv 0\pmod 3$
Do đó pt vô nghiệm.

gt&#x21D2;x2&#x2212;xy&#x2212;(5x&#x2212;5y)&#x2212;x+8=0&#x21D2;(x&#x2212;y)(x&#x2212;5)&#x2212;(x&#x2212;5)=&#x2212;3&#x21D2;(5&#x2212;x)(x&#x2212;y&#x2212;1)=3" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">gt⇒x2−xy−(5x−5y)−x+8=0⇒(x−y)(x−5)−(x−5)=−3⇒(5−x)(x−y−1)=3

3" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">3 là sẽ tìm được nghiệm nguyên của PT