a.

Với \(y=0\) không phải nghiệm

Với \(y\ne0\Rightarrow\left\{{}\begin{matrix}3x+2=\dfrac{5}{y}\\2x\left(x+y\right)+y=\dfrac{5}{y}\end{matrix}\right.\)

\(\Rightarrow3x+2=2x\left(x+y\right)+y\)

\(\Leftrightarrow2x^2+\left(2y-3\right)x+y-2=0\)

\(\Delta=\left(2y-3\right)^2-8\left(y-2\right)=\left(2y-5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-2y+3+2y-5}{4}=-\dfrac{1}{2}\\x=\dfrac{-2y+3-2y+5}{4}=-y+2\end{matrix}\right.\)

Thế vào pt đầu ...

Câu b chắc chắn đề sai

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

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

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

$\Leftrightarrow x^2+2=y$

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

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

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

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

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 ... 

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

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

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

\(\Rightarrow...\)

Lời giải:

Lấy PT(1) trừ đi PT(2) ta thu được:

$x^2+xy-x+y-2y^2=0$

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

$\Leftrightarrow (x-y)(x+y)+y(x-y)-(x-y)=0$

$\Leftrightarrow (x-y)(x+2y-1)=0$

$\Rightarrow x-y=0$ hoặc $x+2y-1=0$

Nếu $x-y=0\Rightarrow x=y$

Thay vào PT(1): $2y^2+3y^2+2y+y=0$

$\Leftrightarrow y=0$ hoặc $y=-\frac{3}{5}$

$y=0$ thì $x=0$

$y=-\frac{3}{5}$ thì $x=\frac{-3}{5}$

Nếu $x+2y-1=0\Rightarrow 2y=1-x$. Thay vào PT(2):

$2x^2+2x(1-x)+(1-x)^2+6x=0$

$\Leftrightarrow x^2+6x+1=0$

$\Rightarrow x=-3\pm 2\sqrt{2}\Rightarrow y=2\mp \sqrt{2}$

Vậy.......

Lời giải:

Lấy PT(1) trừ đi PT(2) ta thu được:

$x^2+xy-x+y-2y^2=0$

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

$\Leftrightarrow (x-y)(x+y)+y(x-y)-(x-y)=0$

$\Leftrightarrow (x-y)(x+2y-1)=0$

$\Rightarrow x-y=0$ hoặc $x+2y-1=0$

Nếu $x-y=0\Rightarrow x=y$

Thay vào PT(1): $2y^2+3y^2+2y+y=0$

$\Leftrightarrow y=0$ hoặc $y=-\frac{3}{5}$

$y=0$ thì $x=0$

$y=-\frac{3}{5}$ thì $x=\frac{-3}{5}$

Nếu $x+2y-1=0\Rightarrow 2y=1-x$. Thay vào PT(2):

$2x^2+2x(1-x)+(1-x)^2+6x=0$

$\Leftrightarrow x^2+6x+1=0$

$\Rightarrow x=-3\pm 2\sqrt{2}\Rightarrow y=2\mp \sqrt{2}$

Vậy.......

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:

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

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

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

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

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

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

\(\Leftrightarrow...\)

a/ \(\Leftrightarrow\left\{{}\begin{matrix}4x^2-16xy+4y^2=4\\y^2-3xy=4\end{matrix}\right.\)

\(\Rightarrow4x^2-13xy+3y^2=0\)

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

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

Thay vào pt sau: \(\left[{}\begin{matrix}y^2-3y.y=4\left(vn\right)\\\left(4x\right)^2-3x.4x=4\end{matrix}\right.\)

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

b/

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

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

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

Thay vào pt đầu: \(\left[{}\begin{matrix}2\left(2y\right)^2-3.2y.y+y^2=3\\2\left(\frac{2}{3}y\right)^2-3.\frac{2}{3}y.y+y^2=3\end{matrix}\right.\) bạn tự giải nốt