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:

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

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

a.

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

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

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

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

Thế vào pt đầu:

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

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

\(\Leftrightarrow...\)

b.

\(x^2-2xy+x=-y\)

Thế vào \(y^2\) ở pt dưới:

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

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

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

\(\left(1\right)\Leftrightarrow2x^2-4xy+2x+4y^2-8y+4=0\)

\(\Leftrightarrow2\left(x^2-2xy+x\right)+4y^2-8y+4=0\)

\(\Leftrightarrow-2y+4y^2-8y+4=0\)

\(\Leftrightarrow...\)

Câu a pt đầu là \(x^2+2xy^2=3\) hay \(x^3+2xy^2=3\) vậy nhỉ? Nhìn \(x^2\) chẳng hợp lý chút nào

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

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

Trừ vế cho vế:

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

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

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

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

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

- Với \(y=x^2\) thế xuống pt dưới:

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

....

Hai trường hợp còn lại bạn tự thế tương tự

Câu 1:

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

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

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

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

\(\Leftrightarrow x-1=y+2\Rightarrow x=y+3\)

Thay vào pt dưới:

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

\(\Leftrightarrow2y^2+9y+6=0\) \(\Rightarrow...\)

Câu 2:

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

\(\Leftrightarrow x^2+4xy+4y^2+3x+6y+2=0\)

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

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

TH1: \(x+2y=-1\Rightarrow x=-2y-1\) thay vào pt dưới:

\(\left(-2y-1\right)y+y^2+3y+1=0\)

\(\Leftrightarrow-y^2+2y+1=0\Rightarrow...\)

TH2: \(x+2y=-2\Rightarrow x=-2y-2\) thay vào pt dưới:

\(\left(-2y-2\right)y+y^2+3y+1=0\)

\(\Leftrightarrow-y^2-y+1=0\Rightarrow...\)

Giải giúp mik câu c thôi cx đc!

Help me !!!