\(\left\{{}\begin{matrix}x-y+1=0\\2x^2-xy+3y^2-7x-12y+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\2\left(y-1\right)^2-\left(y-1\right)y+3y^2-7\left(y-1\right)-12y+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\4y^2-22y+10=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\2y^2-11y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\\left(2y^2-10y\right)-\left(y-5\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\2y\left(y-5\right)-\left(y-5\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\\left(y-5\right)\left(2y-1\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y-1\\y-5=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=y-1\\2y-1=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=\frac{-1}{2}\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

Vậy hpt đã cho có 2 nghiệm (x,y) \(\in\left\{\left(4;5\right),\left(\frac{-1}{2};\frac{1}{2}\right)\right\}\)

\(\left\{{}\begin{matrix}x-y+1=0\\2x^2-xy+3y^2-7x-12y+1=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=y-1\\2\left(y-1\right)^2-\left(y-1\right)y+3y^2-7\left(y-1\right)-12y+1=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=y-1\\2y^2-4y+2-y^2+y+3y^2-7y+7-12y+1=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=y-1\\4y^2-22y+10=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\4y^2-20y-2y+10\end{matrix}\right.\)

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

\(\left\{{}\begin{matrix}x=y-1\\\Leftrightarrow\left[{}\begin{matrix}y-5=0\\2y-1=0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=5-1=4\\x=\frac{1}{2}-1=-\frac{1}{2}\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}y=5\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là : (a, b) ∈ {4, 5; -1/2, 1/2}

\(\left\{{}\begin{matrix}2x^2-xy+3y^2-7x-12y+1=0\left(1\right)\\x-y+1=0\left(2\right)\end{matrix}\right.\)

Từ (2) SUY RA : \(x=y-1\)

Thay x = y - 1 vào (1) được :

\(2\left(y-1\right)^2-\left(y-1\right)y+3y^2-7\left(y-1\right)-12y+1=0\Leftrightarrow2y^2-4y+2-y^2+y+3y^2-7y+7-12y+1=0\Leftrightarrow4y^2-22y+10=0\) \(\Leftrightarrow\) \(\left(y-5\right)\left(2y-1\right)=0\Leftrightarrow\left[{}\begin{matrix}y=5\\y=\dfrac{1}{2}\end{matrix}\right.\)

Suy ra x = 4 hoặc x = -1/2

Vậy nghiệm của hệ pt là (4;5) (-1/2;1/2)

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

\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4\\6x+9y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11y=-14\\3x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{14}{11}\\x=\dfrac{y+2}{3}=\dfrac{\dfrac{14}{11}+2}{3}=\dfrac{12}{11}\end{matrix}\right.\)

\(9,\Leftrightarrow\left\{{}\begin{matrix}3x-2=y\\2x+3\left(3x-2\right)=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2=y\\11x=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{12}{11}\\y=\dfrac{14}{11}\end{matrix}\right.\)

\(10,\Leftrightarrow\left\{{}\begin{matrix}2x=2-3y\\2\left(2-3y\right)-y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2-3y\\4-6y-y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{14}\\y=\dfrac{3}{7}\end{matrix}\right.\)

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

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}y=x^2\\xy+1=x^2-y\end{matrix}\right.\) thay xuống pt dưới:

- Với \(y=x^2\) thay xuống pt dưới \(\Rightarrow x^3=1\)

- Với \(xy+1=x^2-y\) thay xuống dưới:

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

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

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:

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

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

b: \(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x-3y-3=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2+y^2-2x-2y-23=0\\x=3y+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(3y+3\right)^2+y^2-2\left(3y+3\right)-2y-23=0\\x=3y+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9y^2+18y+9+y^2-6y-6-2y-23=0\\x=3y+3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}10y^2+10y-20=0\\x=3y+3\end{matrix}\right.\)

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

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

a: \(\left\{{}\begin{matrix}3x^2+6xy-x+3y=0\\4x-9y=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}9y=4x-6\\3x^2+6xy-x+3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{9}x-\dfrac{2}{3}\\3x^2+6x\cdot\left(\dfrac{4}{9}x-\dfrac{2}{3}\right)-x+3\cdot\left(\dfrac{4}{9}x-\dfrac{2}{3}\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x^2+\dfrac{8}{3}x^2-4x-x+\dfrac{4}{3}x-2=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{17}{3}x^2-\dfrac{11}{3}x-2=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}17x^2-11x-6=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x-1\right)\left(17x+6\right)=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}17x+6=0\\y=\dfrac{4}{9}x-\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=\dfrac{4}{9}\cdot1-\dfrac{2}{3}=\dfrac{4}{9}-\dfrac{2}{3}=-\dfrac{2}{9}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{6}{17}\\y=\dfrac{4}{9}\cdot\dfrac{-6}{17}-\dfrac{2}{3}=\dfrac{-14}{17}\end{matrix}\right.\end{matrix}\right.\)

1) Cộng vế theo vế ta được

\(2x^2+3xy+y^2-7x-5y+6=0\)

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

Thay vào phương trình giải bình thường

2) Nhận thấy \(y=0\)không là nghiệm của hpt trên.Vì thế nhân cả 2 vế của (2) cho 18y ta được:\(72x^2y^{2}+108xy=18y^3\) (3)
Lấy (1) trừ (3) ta được:\(8x^3y^3-72x^2y^{2}-108xy+27=0 \)
Đến đây đặt \(a=xy\) giải bình thường

a, #Góp ý từ nhiều người nhưng họ không giải nên t làm giùm

ĐK: \(x\le3\)

\(\left\{{}\begin{matrix}x^2+y^2+1=2\left(xy-x+y\right)\left(1\right)\\x^3+3y^2+5x-12=\left(12-y\right)\sqrt{3-x}\left(2\right)\end{matrix}\right.\)

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

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

\(\left(2\right)\)\(\Leftrightarrow x^3+3\left(x+1\right)^2+5x-12=\left[12-\left(x+1\right)\right]\sqrt{3-x}\)

\(\Leftrightarrow x^3+3x^2+11x-9=\left(11-x\right)\sqrt{3-x}\)

\(\Leftrightarrow x^3+3x^2+8x=\left(11-x\right)\sqrt{3-x}+3\left(3-x\right)\)

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

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

\(\Leftrightarrow x=\sqrt{3-x}\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+x-3=0\end{matrix}\right.\) \(\Rightarrow x=\frac{-1+\sqrt{13}}{2}\left(tm\right)\Rightarrow y=\frac{1+\sqrt{13}}{2}\)

Vậy...

Akai Haruma, No choice teen, Arakawa Whiter, Phạm Hoàng Lê Nguyên, Vũ Minh Tuấn, tth, HISINOMA KINIMADO, Nguyễn Việt Lâm

Mn giúp e vs ạ! thanks!

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