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

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

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

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?

Viết lại (2)

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

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

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

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

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

- TH1: \(xy=1\)

\(\left(1\right)\Rightarrow5x-4y+3y^3-2\left(x+y\right)=0\)

\(\Leftrightarrow3x-6y+3y^3=0\)

\(\Leftrightarrow\dfrac{3}{y}-6y+3y^3=0\)

Đến đây dễ rồi nhé.

- TH2: \(x^2+y^2=2\)

\(\left(1\right)\Rightarrow5x^2y-4xy^2+3y^3-\left(x^2+y^2\right)\left(x+y\right)=0\)

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

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

Đến đây dễ rồi nhé.

\(\left\{{}\begin{matrix}x^2+y^2=2\left(1\right)\\3y^2+4xy+x+2y=10\left(2\right)\end{matrix}\right.\)

Lấy \(\left(1\right)+\left(2\right)\Leftrightarrow x^2+4xy+4y^2+x+2y=12\)

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

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

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

Với \(x=3-2y\) :

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

\(\Leftrightarrow5y^2-12y+7=0\)

\(\Leftrightarrow\left(y-1\right)\left(5y-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=\frac{7}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{5}\end{matrix}\right.\)

Với \(x=-2y-4\) :

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

\(\Leftrightarrow5y^2+16y+14=0\)

\(\Delta'=8-60=-62< 0\)

\(\Rightarrow PTVN\)

Vậy \(\left[{}\begin{matrix}\left(x;y\right)=\left(1;1\right)\\\left(x;y\right)=\left(\frac{1}{5};\frac{7}{5}\right)\end{matrix}\right.\)

Gọi pt đầu là (1); pt sau là (2).

(2)\(\Leftrightarrow3y^2+\left(4x+2\right)y+x-10=0\)

Coi đây là pt bậc 2 ẩn y với x là tham số.

\(\Delta=\left(4x+2\right)^2-12\left(x-10\right)\)

\(=16x^2+4x+124>0\forall x\)

Pt có 2 ng₀ pb:

\(y_1=\frac{-4x-2+\sqrt{16x^2+4x+124}}{6}\);\(y_2=\frac{-4x-2-\sqrt{16x^2+4x+124}}{6}\)

-Xét y₁:

Thay vào (1):\(x^2+\frac{\left[\sqrt{16x^2+4x+124}-\left(4x+2\right)\right]^2}{36}-2=0\)

\(\Leftrightarrow64x^2+20x+126=\left(16x+8\right)\sqrt{4x^2+x+31}\)(Ở bước này bạn nhân với 36 rồi biến đổi cho gọn).

Đến đây dùng máy tính giải hoặc bình phương lên rồi giải.

Làm ttự với y₂ để tìm x,y.

Nguyễn Việt Lâm Nhờ bn làm cách khác gọn hơn.

ý a ở đây bn https://hoc247.net/hoi-dap/toan-10/giai-he-pt-3x-x-2-2-y-2-va-3y-y-2-2-x-2-faq371128.html

b.

Với \(xy=0\) không là nghiệm

Với \(xy\ne0\)

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

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

\(\Rightarrow\dfrac{5-x^2}{x}=5-2x\)

\(\Leftrightarrow5-x^2=5x-2x^2\)

\(\Leftrightarrow...\)

c.

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

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

\(\Rightarrow5x^2-x\left(y+1\right)-4\left(y+1\right)^2=0\)

\(\Leftrightarrow\left(x-y-1\right)\left(5x+4\left(y+1\right)\right)=0\)

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

Thế vào 1 trong 2 pt ban đầu...

a: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\cdot\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=1+2=3\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}3x+2y=2\\6x-3y=18\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=2-2y\\2\cdot3x-3y=18\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=2-2y\\2\left(2-2y\right)-3y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4-7y=18\\3x=2-2y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7y=-14\\3x=2-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\3x=2-2\cdot\left(-2\right)=6\end{matrix}\right.\)

=>x=2 và y=-2