1.

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

\(\Leftrightarrow \left\{\begin{matrix} -x+y=5\\ 2x-4y=9\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x=\frac{-29}{2}\\ y=\frac{-19}{2}\end{matrix}\right.\)

Vậy.............

2.

ĐKXĐ: $x\in\mathbb{R}$

$x^2+x-2\sqrt{x^2+x+1}+2=0$

$\Leftrightarrow (x^2+x+1)-2\sqrt{x^2+x+1}+1=0$

$\Leftrightarrow (\sqrt{x^2+x+1}-1)^2=0$

$\Rightarrow \sqrt{x^2+x+1}=1$

$\Rightarrow x^2+x=0$

$\Leftrightarrow x(x+1)=0$

$\Rightarrow x=0$ hoặc $x=-1$

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...

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

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

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

$\Leftrightarrow x^2+2=y$

\(\hept{\begin{cases}x+y-2xy=0\\x+y-x^2y^2=\sqrt{\left(xy-1\right)^2+1}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+y=2xy\\2xy-x^2y^2=\sqrt{x^2y^2-2xy+2}\left(1\right)\end{cases}}\)

đặt 2xy-x^2y^2=t 

=> (1)  \(\Leftrightarrow t=\sqrt{2-t}\)

Tự làm nốt nhé

\(\Leftrightarrow\hept{\begin{cases}x+y-2xy=0\\x+y-x^2y^2=\sqrt{x^2y^2-2xy+2}\end{cases}}\)

Đặt x+y=a, xy=b

\(\Rightarrow\hept{\begin{cases}a-2b=0\\a-b^2=\sqrt{b^2-2b+2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=2b\\2b-b^2=\sqrt{b^2-2b+2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=2b\\b^4-4b^3+4b^2=b^2-2b+2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=2b\\b^4-4b^3+3b^2+2b-2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=2b\\\left(b-1\right)^2\left(b^2-2b-2\right)=0\end{cases}}\)

Đến đây đơn giản rồi nhé :P