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:

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

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

\(\left\{{}\begin{matrix}x^3-3x^2-9x+22=y^3+3y^2-9y\left(1\right)\\x^2+y^2-x+y=\dfrac{1}{2}\left(2\right)\end{matrix}\right.\)

PT (1)\(\Leftrightarrow\left(x-y\right)^3+3xy\left(x-y\right)-3\left(x^2+y^2\right)-9\left(x-y\right)=-22\)

\(\Leftrightarrow\left(x-y\right)^3+3xy\left(x-y\right)-3\left(x-y\right)^2-6xy-9\left(x-y\right)=-22\)

PT (2)\(\Leftrightarrow\left(x-y\right)^2-\left(x-y\right)+2xy=\dfrac{1}{2}\)

Đặt \(\left\{{}\begin{matrix}a=x-y\\b=xy\end{matrix}\right.\)

Hệ tt \(\left\{{}\begin{matrix}a^3+3ab-3a^2-6b-9a=-22\\a^2-a+2b=\dfrac{1}{2}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}a^3+3ab-3a^2-6b-9a=-22\\b=\dfrac{1-2a^2+2a}{4}\end{matrix}\right.\)

\(\Rightarrow a^3+3a\left(\dfrac{1-2a^2+2a}{4}\right)-3a^2-6\left(\dfrac{1-2a^2+2a}{4}\right)-9a=-22\)

\(\Leftrightarrow-2a^3+6a^2-45a+82=0\)

\(\Leftrightarrow a=2\)\(\Rightarrow b=-\dfrac{3}{4}\)

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

Vậy...

pt1:

(x-1)^3-12x+12=(y+1)^3-12y-12

<=> (x-1)^3 -12(x-1)=(y+1)^3-12(y+1). đặt x-1=a ; y+1=b

a^3-12a=b^3-12b

=>(a-b) (a^2+ab+b^2-12)=0

với a^2+b^2+ab=12

(x-1)^2+(y+1)^2+(x-1)(y+1)=12

x^2+y^2+xy-x+y=11(1)

kết hợp pt2 x^2+y^2+y-x=1/2 thay vào (1)

xy=21/2 từ đây thế x theo y dễ dàng giải nghiệm

với a=b thì x-1=y+1

=>từ đây thế x theo y vào pt2 ta dễ dàng giải

a)

đặt \(x^2-x=u;y^2-2y=v\)

hpt trở thành

\(\left\{{}\begin{matrix}u+v=19\\uv=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}u=\dfrac{19-\sqrt{281}}{2}\\v=\dfrac{19+\sqrt{281}}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}u=\dfrac{19+\sqrt{281}}{2}\\v=\dfrac{19-\sqrt{281}}{2}\end{matrix}\right.\end{matrix}\right.\)

dễ thấy tại 2 trường hợp hpt đều vô nó nên hpt vô no

đc 1 câu

Hiếu Cao Huy

a.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\5\sqrt{x-2}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\\sqrt{x-2}=1\end{matrix}\right.\)

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

b.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)

a.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)

b.

ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)

Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:

\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)

\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)

\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)

Thay xuống pt dưới:

\(6y+y=14\Rightarrow y=2\)

\(\Rightarrow x=4\)

a: =>xy-2x+2y-4=xy+y và 5xy+10x+y+2=5xy-10x-2y+4

=>-2x+y=4 và 20x+3y=2

=>x=-5/13; y=42/13

b: =>4x+2|y|=8 và 4x-3y=1

=>2|y|-3y=7 và 4x-3y=1

TH1: y>=0

=>2y-3y=7 và 4x-3y=1

=>-y=7 và 4x-3y=1

=>y=-7(loại)

TH2: y<0

=>-2y-3y=7 và 4x-3y=1

=>y=-7/5; 4x=1+3y=1-21/5=-16/5

=>x=-4/5; y=-7/5