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

\(\Delta=\left(3y-3\right)^2-8\left(y^2-1y+1\right)=\left(y-1\right)^2\)

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

\(\Rightarrow...\)

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:

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

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

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

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

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

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

\(\Leftrightarrow...\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-6xy+2y^2=6\\x^2+2xy-2y^2=6\end{matrix}\right.\)

\(\Rightarrow3x^2-8xy+4y^2=0\)

\(\Rightarrow\left(3x-2y\right)\left(x-2y\right)=0\)

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

Thế vào pt đầu...

\(\left\{{}\begin{matrix}2x^2-3xy+y^2=3\\x^2+2xy-2y^2=6\end{matrix}\right.\)\(\left(1\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-6xy+2y^2=6\\x^2+2xy-2y^2=6\end{matrix}\right.\)

\(\Leftrightarrow3x^2-8xy+4y^2=0\)

\(\Leftrightarrow3x\left(x-2y\right)-2y\left(x-2y\right)=0\)

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

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

Thay vào \(\left(1\right)\) ta được:

\(\Leftrightarrow\left[{}\begin{matrix}2.\left(2y\right)^2-3.2y.y+y^2=3\\2.\left(\dfrac{2y}{3}\right)^2-3.\dfrac{2y}{3}.y+y^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}y^2=1\\y^2=-27\left(VLý\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\end{matrix}\right.\)

Vậy ... 

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:

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

\(\Rightarrow x=4\)

b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)

\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)

\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)

\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)

\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)

caau a) binh phuong len ra no x=y tuong tu

c)

ĐK $y \geqslant 0$

Hệ đã cho tương đương với

$\left\{\begin{matrix} 2x^2+2xy+2x+6=0\\ (x+1)^2+3(y+1)+2xy=2\sqrt{y(x^2+2)} \end{matrix}\right.$

Trừ từng vế $2$ phương trình ta được

$x^2+2+2\sqrt{y(x^2+2)}-3y=0$

$\Leftrightarrow (\sqrt{x^2+2}-\sqrt{y})(\sqrt{x^2+2}+3\sqrt{y})=0$

$\Leftrightarrow x^2+2=y$

1. \(\left\{{}\begin{matrix}3x+4y=11\\2x-y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+4y=11\\8x-4y=-44\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+4y=11\\11x=-33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=-3\end{matrix}\right.\)

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

3.\(\left\{{}\begin{matrix}3x+\dfrac{5}{2}y=9\\2x+\dfrac{1}{3}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+5y=18\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y=12\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Điều kiện: \(y\ge0\)

pt thứ nhất của hệ \(\Leftrightarrow\left(y-x+3\right)^2=0\) \(\Leftrightarrow y-x+3=0\) \(\Leftrightarrow y=x-3\)

Thay vào pt thứ hai của hệ, ta được  \(2x^2+3x+x-3-\left(3x+1\right)\sqrt{x-3}-2=0\)

\(\Leftrightarrow2x^2+4x-5=\left(3x+1\right)\sqrt{x-3}\)         \(\left(x\ge3\right)\)

\(\Rightarrow\left(2x^2+4x-5\right)^2=\left[\left(3x+1\right)\sqrt{x-3}\right]^2\)

\(\Leftrightarrow4x^4+16x^2+25+16x^3-20x^2-40x=\left(3x+1\right)^2\left(x-3\right)\)

\(\Leftrightarrow4x^4+16x^3-4x^2-40x+25=9x^3-21x^2-17x-3\)

\(\Leftrightarrow4x^4+7x^3+17x^2-23x+28=0\)

Đặt \(f\left(x\right)=4x^4+7x^3+17x^2-23x+28\)

\(f\left(x\right)=4x^4+7x^3+17x^2+4+4+...+4-23x+4\) (có 6 số 4 ở giữa)

\(f\left(x\right)\ge9\sqrt[9]{4x^4.7x^3.17x^2.4^6}-23x+4\) \(=\left(9\sqrt[9]{1949696}-23\right)x+4\)

Hiển nhiên \(9\sqrt[9]{1949696}>23\). Lại có \(x\ge3\) nên \(f\left(x\right)>0\), Như vậy pt \(f\left(x\right)=0\) vô nghiệm. Điều đó có nghĩa là phương trình đã cho vô nghiệm.