mấy bài dạng như này mk sẽ hướng dẩn nha .

a) ta có : \(\left\{{}\begin{matrix}\left(x+y-2\right)\left(2x-y\right)=0\\x^2+y^2=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+y-2=0\\2x-y=0\end{matrix}\right.\\x^2+y^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y-2=0\\x^2+y^2=2\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y=0\\x^2+y^2=0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\) giải bằng cách thế bình thường nha

b) ta có : \(\left\{{}\begin{matrix}x^2+y^2+2x+2y=6\\x+y-3xy+1=0\end{matrix}\right.\) \(\Leftrightarrow2x^2+2y^2+6xy-5=0\)

\(\Leftrightarrow2\left(x+y\right)^2+2xy-5=0\) sài vi ét --> .......................

c) đây là phương trình đối xứng loại 1 , có trên mang nha .

câu d và e là phương trình đối xứng loại 2 , cũng có trên mạng nha .

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

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

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

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?

a.

\(\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)

Nhân vế:

\(-4\left(x^3-y^3\right)=\left(16x-4y\right)\left(5x^2-y^2\right)\)

\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)

\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4y}{7}\\y=-3x\end{matrix}\right.\)

Thế vào \(y^2=5x^2+4...\)

b. Đề bài không hợp lý ở \(4x^2\)

c.

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=9\\3x^2+6y^2=3x-12y\end{matrix}\right.\)

Trừ vế:

\(x^3-y^3-3x^2-6y^2=9-3x+12y\)

\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)

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

\(\Leftrightarrow x-1=y+2\)

\(\Leftrightarrow y=x-3\)

Thế vào \(x^2=2y^2=x-4y\) ...

b.

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2+y^4-4xy^3=1\\4x^2+2y^2-4xy=2\end{matrix}\right.\)

\(\Rightarrow y^4-2y^2-4xy^3+4xy=-1\)

\(\Leftrightarrow\left(y^2-1\right)^2-4xy\left(y^2-1\right)=0\)

\(\Leftrightarrow\left(y^2-1\right)\left(y^2-1-4xy\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\\x=\dfrac{y^2-1}{4y}\end{matrix}\right.\)

Thế vào \(2x^2+y^2-2xy=1\) ...

Với \(x=\dfrac{y^2-1}{4y}\) ta được:

\(2\left(\dfrac{y^2-1}{4y}\right)^2+y^2-2\left(\dfrac{y^2-1}{4y}\right)y=1\)

\(\Leftrightarrow5y^4-6y^2+1=0\)

b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)

\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)

\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)

\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)

\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)

caau a) binh phuong len ra no x=y tuong tu

c)

ĐK $y \geqslant 0$

Hệ đã cho tương đương với

$\left\{\begin{matrix} 2x^2+2xy+2x+6=0\\ (x+1)^2+3(y+1)+2xy=2\sqrt{y(x^2+2)} \end{matrix}\right.$

Trừ từng vế $2$ phương trình ta được

$x^2+2+2\sqrt{y(x^2+2)}-3y=0$

$\Leftrightarrow (\sqrt{x^2+2}-\sqrt{y})(\sqrt{x^2+2}+3\sqrt{y})=0$

$\Leftrightarrow x^2+2=y$

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

\(\left(1\right)\Leftrightarrow\left(2x^2+2xy-4x\right)+\left(-xy-y^2+2y\right)+\left(-x-y+2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}y=2-x\left(3\right)\\y=2x-1\left(4\right)\end{matrix}\right.\)

Thế (3) vào (2) ta được:

\(x^3+\left(2-x\right)^2+x+\left(2-x\right)-4=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{3}-1\\x=-\sqrt{3}-1\end{matrix}\right.\)

Tương tự cho trường hợp còn lại.

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

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

*)Xét \(y=2x-1\) thì

\(pt\left(2\right)\Leftrightarrow x^3+\left(2x-1\right)^2+x+\left(2x-1\right)-4=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-4\\x=1\\x=-1\end{matrix}\right.\)

*)Xét \(y=2-x\) thì:

\(pt\left(2\right)\Leftrightarrow x^3+\left(2-x\right)^2+x+\left(2-x\right)-4=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x^2+2x-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-2\pm\sqrt{12}}{2}\end{matrix}\right.\)

\(y^2-\left(x+1\right)y-2x^2+5x+2=0\Leftrightarrow\left(y+x-2+1\right)=0.\)Do đó hệ đã cho tương đương với:

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

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{4}{5}\\y=-\dfrac{13}{5}.\end{matrix}\right.\end{matrix}\right.\)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)

\(\Rightarrow-4\left(x^3-y^3\right)=\left(5x^2-y^2\right)\left(16x-4y\right)\)

\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)

\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\y=\dfrac{7x}{4}\\y=-3x\end{matrix}\right.\)

Lần lượt thế vào \(y^2=5x^2+4\)...

b. Đề bài bất hợp lý, \(4x^2+y^4\) cần là \(4x^4+y^4\)

a.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}y+2=0\left(\text{không thỏa mãn}\right)\\x^2+y^2-4=x\left(x+y-2\right)\end{matrix}\right.\) 

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

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

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

Thế vào pt dưới:

\(\Rightarrow\left[{}\begin{matrix}x^2+8+2x+2x-4=0\\\left(y+2\right)^2+2y^2+y\left(y+2\right)+2\left(y+2\right)-4=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

Câu b chắc chắn đề sai, nhìn 2 vế pt đầu đều có \(x^2\) thì chúng sẽ rút gọn, không ai cho đề như thế hết

1.

ĐKXĐ: ....

\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2x^2-1=xy\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2x-\dfrac{1}{x}=y\end{matrix}\right.\)

Trừ vế cho vế: \(\Rightarrow x=\dfrac{1}{y}\Rightarrow xy=1\)

Thay xuống pt dưới: \(2x^2-2=0\Leftrightarrow x^2=1\Leftrightarrow...\)

2.

Với \(y=0\) không phải nghiệm

Với \(y\ne0\)

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

Cộng vế với vế:

\(4x^3+3x=4\left(\dfrac{1}{y}\right)^3+3\left(\dfrac{1}{y}\right)\)

\(\Leftrightarrow4\left(x^3-\dfrac{1}{y^3}\right)+3\left(x-\dfrac{1}{y}\right)=0\)

\(\Leftrightarrow4\left(x-\dfrac{1}{y}\right)\left(x^2+\dfrac{x}{y}+y^2\right)+3\left(x-\dfrac{1}{y}\right)=0\)

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

\(\Leftrightarrow x-\dfrac{1}{y}=0\Leftrightarrow y=\dfrac{1}{x}\)

Thế vào pt đầu:

\(4x^3+1=3x\)

\(\Leftrightarrow4x^3-3x+1=0\)

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

\(\Leftrightarrow...\)