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...

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

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

Lời giải:

1. Ta thấy: 
$(1-x)^2\geq 0; (3-y)^2\geq 0; (y^2-x-z)^2\geq 0$ với mọi $x,y,z$

Do đó để tổng của chúng bằng $0$ thì $(1-x)^2=(3-y)^2=(y^2-x-z)^2=0$

$\Rightarrow x=1; y=3; z=y^2-x=3^2-1=8$

2.

Bạn xem có viết lộn dấu bình phương ở cụm ( ) thứ nhất vào bên trong không vậy>

\(\Leftrightarrow\left\{{}\begin{matrix}6x-6-2y+4=0\\4x+4-3y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=1\\4x-3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\)  \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)

TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)

Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)

TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\) 

\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)

2 câu dưới hình như em hỏi rồi?

Trước chủ nhật 

=))

a.

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

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

Xét phương trình: \(\left|y-2\right|-y=-1\)

TH1: \(y\ge2\)

\(\Rightarrow y-2-y=-1\Leftrightarrow-2=-1\) (loại)

TH2: \(y\le2\)

\(\Rightarrow2-y-y=-1\Rightarrow y=\dfrac{3}{2}\)

Thế vào \(\left|x-1\right|+y=3\)

\(\Rightarrow\left|x-1\right|+\dfrac{3}{2}=3\Rightarrow\left|x-1\right|=\dfrac{3}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{3}{2}\Rightarrow x=\dfrac{5}{2}\\x-1=-\dfrac{3}{2}\Rightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}\left|x+1\right|+\left|y-1\right|=5\\\left|x+1\right|-4y+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|y-1\right|+4y-4=5\\\left|x+1\right|-4y+4=0\end{matrix}\right.\)

Xét phương trình: \(\left|y-1\right|+4y-4=5\)

TH1: \(y\ge1\)

\(\Rightarrow y-1+4y-4=5\Rightarrow y=2\)

Thế vào \(\left|x+1\right|-4y+4=0\)

\(\Rightarrow\left|x+1\right|=4\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

TH2: \(y\le1\)

\(\Rightarrow1-y+4y-4=5\Rightarrow y=\dfrac{8}{3}>1\) (không thỏa mãn)

toi ko bt

\(x^2+y^2=0\)

Mà \(x^2\ge0;y^2\ge0\)nên \(x^2+y^2\ge0\)

(Dấu "="\(\Leftrightarrow x=y=0\))

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

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

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

Mà \(\left(x-y\right)^2+\left(y+1\right)^2\ge0\)

(Dấu "="\(\Leftrightarrow x=-1;y=-1\)