a.

\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

TH1:

\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

c.

\(\left\{{}\begin{matrix}\left(x+y\right)^2-4\left(x+y\right)-12=0\\\left(x-y\right)^2-2\left(x-y\right)=3\end{matrix}\right.\)

Xét pt:

\(\left(x+y\right)^2-4\left(x+y\right)-12=0\)

\(\Leftrightarrow\left(x+y+2\right)\left(x+y-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y+2=0\\x+y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=-x-2\\y=6-x\end{matrix}\right.\)

TH1: \(y=-x-2\) thế vào \(\left(x-y\right)^2-2\left(x-y\right)=3\)

\(\Rightarrow\left(2x+2\right)^2-2\left(2x+2\right)=3\)

\(\Leftrightarrow4x^2+4x-3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow y=-\dfrac{5}{2}\\x=-\dfrac{3}{2}\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)

TH2: \(y=6-x\) thế vào...

\(\left(2x-6\right)^2-2\left(2x-6\right)=3\)

\(\Leftrightarrow4x^2-28x+45=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\Rightarrow y=\dfrac{7}{2}\\y=\dfrac{9}{2}\Rightarrow y=\dfrac{3}{2}\end{matrix}\right.\)

\(\left(x-3\right)^2+\left(y+2\right)^2=0\)

\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\\\left(y+2\right)^2=0\Rightarrow y+2=0\Rightarrow y=-2\end{matrix}\right.\)

đề sai câu b các câu sau áp dụng tương tự

c/ Vì: \(\left(x-12+y\right)^{200}+\left(x-4-x\right)^{200}=0\)

mà \(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}\ge0\forall x,y\\\left(x-4-y\right)^{200}\ge0\forall x,y\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-12+y\right)^{200}=0\\\left(x-4-y\right)^{200}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-12+y=0\\x-4-y=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=12\\x-y=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=4\end{matrix}\right.\)

Ta thấy \(\left(x+y-z\right)^2\ge0\); \(\left(x-y+2\right)^2\ge0\);\(\left(x+4\right)^2\ge0\)với mọi x,y,z

Suy ra \(\left(x+y-z\right)^2+\left(x-y+2\right)^2+\left(x+4\right)^2\ge0\)với mọi x,y,z

Mặt khác \(\left(x+y-z\right)^2+\left(x-y+2\right)^2+\left(x+4\right)^2=0\)

Nên \(\hept{\begin{cases}x+y-z=0\\x-y+2=0\\x+4=0\end{cases}\Rightarrow\hept{\begin{cases}x+y=z\\x+2=y\\x=-4\end{cases}\Rightarrow}\hept{\begin{cases}x+y=z\\y=-2\\x=-4\end{cases}\Rightarrow}\hept{\begin{cases}z=-6\\y=-2\\x=-4\end{cases}}}\)

Vậy.....

vì \(\left(4x^2-4x+1\right)^{2022}\ge0\left(\forall x\right)\),\(\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}\ge0\left(\forall y\right)\),\(\left|x+y+z\right|\ge0\)

mà \(\left(4x^2-4x+1\right)^{2022}+\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)

=>\(\left\{{}\begin{matrix}4x^2-4x+1=0\\y^2+\dfrac{4}{5}y+\dfrac{4}{25}=0\\x+y-z=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-1=0\\y+\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\\dfrac{1}{2}-\dfrac{2}{5}-z=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)

KL: vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)

(x-12+y)² + (y+4-x)²=0

x-12+y=0 => x+y=12

y+4-x=0 =>y-x=-4

=>x=8 , y=4

vì (x-12+y)² >0:(y+4-x)² >0

=>x-12+y=y+4-x=0

=>x-12+x=y+4-y

=>2x-12=4

=>x=8 hoặc -8

thay x vào là ra y nha.

=> \(\hept{\begin{cases}x^2+2xy+y^2-4x+4y=12\\x^2-2xy+y^2-2x-2y=3\end{cases}}\)

 Rồi đến đây tự làm nhé

HPT <=> \(\hept{\begin{cases}\left(x+y\right)^2-4\left(x+y\right)+4=16\\\left(x-y\right)^2-2\left(x-y\right)+1=4\end{cases}}\)<=> \(\hept{\begin{cases}\left(x+y-2\right)^2=4^2\\\left(x-y-1\right)^2=2^2\end{cases}}\)

=> \(\hept{\begin{cases}x+y-2=\pm4\\x-y-1=\pm2\end{cases}}\)

Có các TH:

1/ \(\hept{\begin{cases}x+y-2=4\\x-y-1=2\end{cases}}\)=> \(\hept{\begin{cases}x+y=6\\x-y=3\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{9}{2}\\y=\frac{3}{2}\end{cases}}\)

2/ \(\hept{\begin{cases}x+y-2=4\\x-y-1=-2\end{cases}}\)=> \(\hept{\begin{cases}x+y=6\\x-y=-1\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{5}{2}\\y=\frac{7}{2}\end{cases}}\)

3/ \(\hept{\begin{cases}x+y-2=-4\\x-y-1=2\end{cases}}\)=> \(\hept{\begin{cases}x+y=-2\\x-y=3\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{5}{2}\end{cases}}\)

4/ \(\hept{\begin{cases}x+y-2=-4\\x-y-1=-2\end{cases}}\)=> \(\hept{\begin{cases}x+y=-2\\x-y=-1\end{cases}}\)=> \(\hept{\begin{cases}x=-\frac{3}{2}\\y=-\frac{1}{2}\end{cases}}\)

b: Ta có: \(\left(4x^4-3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)

\(\Leftrightarrow-4x+3+5x+2=0\)

\(\Leftrightarrow x=-5\)