Giải hệ phương trình $\left\{\begin{array}{l}2x^2-y^2+xy-5x+y+2=\sqrt{y-2x+1}-\sqrt{3-3x}\\ x^2-y-1=\sqrt{4x+y+5}-\sqrt{x+2y-2}\end{array}\right.$

Ta có (1) $\iff x^4+x^2+20=y^2+y$

Ta thấy: $$\begin{gathered} x^4+x^2<x^4+x^2+20\le x^4+x^2+20+8x^2 \\ \iff x^2(x^2+1)<y(y+1)\le (x^2+4)(x^2+5) \end{gathered}$$

Vì x, y ∈ Z nên ta xét các trường hợp sau

+ TH1. $$\begin{gathered} y(y+1)=(x^2+1)(x^2+2)\iff x^4+x^2+20=x^4+3x^2+2 \\ \iff 2x^2=18\iff x^2=9\iff x=\pm 3 \end{gathered}$$

Với $x^2=9 \Rightarrow y^2+y=9^2+9+20\iff y^2+y-110=0\iff y=10;y=-11(t.m)$

+ TH2 $$\begin{gathered} y(y+1)=(x^2+2)(x^2+3)\iff x^4+x^2+20=x^4+5x^2+6 \\ \iff 4x^2=14\iff x^2=\frac{7}{2} \left(loại\right) \end{gathered}$$

+ TH3: $y(y+1)=(x^2+3)(x^2+4)\iff 6x^2=8\iff x^2=\frac{4}{3} \left(loại\right)$

+ TH4: $y(y+1)=(x^2+4)(x^2+5)\iff 8x^2=0\iff x^2=0\iff x=0$

Với $x^2=0$ ta có $y^2+y=20\iff y^2+y-20=0\iff y=-5;y=4$

Vậy PT đã cho có nghiệm nguyên (x;y) là :

(3;10), (3;-11), (-3; 10), (-3;-11), (0; -5), (0;4).

$\begin{array}{l}xy+\sqrt{(1+x^2)(1+y^2)}=1\iff \sqrt{{(1+x)}^2{(1+y)}^2}=1-xy\\ \Rightarrow (1+x^2)(1+y^2)={\left(1-xy\right)}^2\\ \iff 1+x^2+y^2+x^2{y}^2=1-2xy+x^2{y}^2\\ \iff x^2+y^2+2xy=0\iff {\left(x+y\right)}^2=0\iff y=-x\\ \Rightarrow x\sqrt{1+y^2}+y\sqrt{1+x^2}=x\sqrt{1+x^2}-x\sqrt{1+x^2}=0\end{array}$