Cộng vế với vế:

\(x^2+2xy+y^2+x+y=12\)

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

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

TH1: \(\left\{{}\begin{matrix}x+y=-4\\xy=5-\left(x+y\right)=9\end{matrix}\right.\)

Theo Viet đảo, x và y là nghiệm: \(t^2-4t+9=0\) (vô nghiệm)

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

Theo Viet đảo, x và y là nghiệm:

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

\(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)

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

a) x + y = 6 (1)

2x - 3y = 12 (2)

(1) ⇔ x = 6 - y (3)

Thế (3) vào (2) ta có:

2(6 - y) - 3y = 12

⇔ 12 - 2y - 3y = 12

⇔ -5y = 12 - 12

⇔ -5y = 0

⇔ y = 0

Thế y = 0 vào (3) ta có:

x = 6 - 0

⇔ x = 6

Vậy S = {6; 0}

b) x - y = 5  (4)

(x - 2)(y + 3) = 3 + xy (5)

(5) ⇔ xy + 3x - 2y - 6 = 3 + xy

⇔ 3x - 2y = 3 + 6

⇔ 3x - 2y = 9 (6)

(4) ⇔ x = y + 5 (7)

Thế x = y + 5 vào (6) ta có:

(6) ⇔ 3(y + 5) - 2y = 9

⇔ 3y + 15 - 2y = 9

⇔ y = 9 - 15

⇔ y = -6

Thế y = -6 vào (7) ta có:

x = -6 + 5

⇔ x = -1

Vậy S ={-1; -6}

Xét \(y=0\)\(\Rightarrow...\)

Xét \(y\ne0\). Ta có:

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

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

Thay (1) vào (2), ta có:

\(\left(5y-y^2-xy\right)\left(x+y-3\right)=-3y\)

\(-y\left(x+y-5\right)\left(x+y-3\right)=-3y\)

\(\Leftrightarrow\left(x+y-5\right)\left(x+y-3\right)=3\left(\cdot\right)\)

Đặt \(x+y-5=t\), phương trình \(\left(\cdot\right)\) trở thành

\(t\left(t+2\right)=3\)\(\Leftrightarrow t^2+2t+1=4\Leftrightarrow\left(t+1\right)^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}t+1=2\\t+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-3\end{matrix}\right.\)

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

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

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)

\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3+7\left(xy+x+y+1\right)=31\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3+\left(xy\right)^3+7\left(xy+x+y\right)=30\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)

\(\Rightarrow\left\{{}\begin{matrix}uv=2\\u^3+v^3+7\left(u+v\right)=30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3-3uv\left(u+v\right)+7\left(u+v\right)=30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3+\left(u+v\right)-30=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)

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

2.

ĐKXĐ: \(0\le x\le\dfrac{3}{2}\)

\(\Leftrightarrow9x\left(3-2x\right)+81+54\sqrt{x\left(3-2x\right)}=49x+25\left(3-2x\right)+70\sqrt{x\left(3-2x\right)}\)

\(\Leftrightarrow9x^2-14x-3+8\sqrt{x\left(3-2x\right)}=0\)

\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(3-x-2\sqrt{x\left(3-2x\right)}\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2-\dfrac{36\left(x-1\right)^2}{3-x+2\sqrt{x\left(3-2x\right)}}=0\)

\(\Leftrightarrow9\left(x-1\right)^2\left(1-\dfrac{4}{3-x+2\sqrt{x\left(3-2x\right)}}\right)=0\)

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

\(\left(1\right)\Leftrightarrow2\sqrt{x\left(3-2x\right)}=x+1\)

\(\Leftrightarrow4x\left(3-2x\right)=x^2+2x+1\)

\(\Leftrightarrow9x^2-10x+1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)