$(x-1)^2+(y+3)^2=0$

Vì $(x-1)^2\ge0$ và $(y+3)^2\ge0$ nên:

$(x-1)^2=0$

$(y+3)^2=0$

$x-1=0$

$y+3=0$

$x=1,\ y=-3$

Vậy $(x,y)=(1,-3)$.

$2(x-5)^4+5|2y-7|^5=0$

Vì $2(x-5)^4\ge0$ và $5|2y-7|^5\ge0$ nên:

$2(x-5)^4=0$

$5|2y-7|^5=0$

$(x-5)^4=0$

$|2y-7|^5=0$

$x-5=0$

$2y-7=0$

$x=5,\ y=\dfrac{7}{2}$

Vậy $(x,y)=\left(5,\dfrac{7}{2}\right)$.

$3(x-2y)^{2004}+4\left|y+\dfrac{1}{2}\right|=0$

Vì $3(x-2y)^{2004}\ge0$ và $4\left|y+\dfrac{1}{2}\right|\ge0$ nên:

$3(x-2y)^{2004}=0$

$4\left|y+\dfrac{1}{2}\right|=0$

$x-2y=0$

$y+\dfrac{1}{2}=0$

$y=-\dfrac{1}{2}$

$x=2y=-1$

Vậy $(x,y)=\left(-1,-\dfrac{1}{2}\right)$.

$|x+3y-1|+\left(2y-\dfrac{1}{2}\right)^{2000}=0$

Vì $|x+3y-1|\ge0$ và $\left(2y-\dfrac{1}{2}\right)^{2000}\ge0$ nên:

$|x+3y-1|=0$

$\left(2y-\dfrac{1}{2}\right)^{2000}=0$

$x+3y-1=0$

$2y-\dfrac{1}{2}=0$

$y=\dfrac{1}{4}$

$x+3\cdot\dfrac{1}{4}=1$

$x=\dfrac{1}{4}$

Vậy $(x,y)=\left(\dfrac{1}{4},\dfrac{1}{4}\right)$.

x-2y+y=0

x-3y=0

3y=x

Thay 3y=x vào x-2y+y=0 ta có:

3y-2y+y=0

y+y=0

2y=0

y=0

mà 3y=x =>x=3.0=0

Vậy x=0 và y=0 thì x-2y+y=0

x - 2y + y = 0

x - y = 0

=> x = y

=>x = y và x , y thuộc Z = {.... ; -3 ; -2 ; -1 ; 0 ; 1 ; 2 ; 3 ; ....}

m

2024 thì bn này ngoài 20 tuổi r trả lời gì nữa :>

\(xy+x-2y+1=0\Leftrightarrow x\left(y+1\right)-2y+2+1=2\Leftrightarrow\left(x+2\right)\left(y+1\right)=1;\)

\(1=1.1=-1.-1\)

\(+,\left(x+2\right)\left(y+1\right)=1.1\Leftrightarrow x=-1;y=0\)

\(+,\left(x+2\right)\left(y+1\right)=-1.-1\Leftrightarrow x=-3;y=-2\)

Đề \(\Leftrightarrow x^2-2xy+y^2+y^2+2y+1+x^2+2x+1-x^2+2x-1+12=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y+1\right)^2+\left(x+1\right)^2-\left(x-1\right)^2+12=0\left(1\right)\)

Ta có: \(\left(x-y\right)^2\ge0,\left(y+1\right)^2\ge0,\left(x+1\right)^2\ge0\ge-\left(x-1\right)^2\)

nên \(\left(x-y\right)^2+\left(y+1\right)^2+\left(x+1\right)^2-\left(x-1\right)^2>0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y+1\right)^2+\left(x+1\right)^2-\left(x-1\right)^2+12>12>0\)

\(\Rightarrow\left(1\right)\)vô lí.

Vậy \(S=\varnothing\)