Giải hệ phương trình: $$\begin{cases}\left(x-y+6\right)\sqrt{x-y+2}-\left(y+7\right)\sqrt{y+3}=3\left(x-2y-1\right)\\2\sqrt{4y+1}+\sqrt{x^3-5x+14y}=\sqrt{3x^3+17x+2y-26}\end{cases}$$
Đặt $\sqrt{x-y+2}=a\geq 0= > x-y+6=a^2+4$
$\sqrt{y+3}\geq 0= > y+7=b^2+4$
$= > x-2y-1=a^2-b^2$
Thay vào đề bài
PT (1) $< = > a(a^2+4)-b(b^2+4)=3(a^2-b^2)< = > (a-b)(a^2+ab+b^2-3a-3b+4)=0$
-Nếu $a-b=0= > a=b= > \sqrt{x-y+2}=\sqrt{y+3}= > 2y=x-1$. Thay vào pt(2)
$= > 2\sqrt{2x-1}+\sqrt{x^3+2x-7}=\sqrt{3x^3+18x-27}= > 4(2x-1)+x^3+2x-7+4\sqrt{(2x-1)(x^3+2x-7)}=3x^3+18x-27= > 2\sqrt{(2x-1)(x^3+2x-7)}=x^3+4x-8=(x^3+2x-7)+(2x-1)= > (\sqrt{x^3+2x-7}-\sqrt{2x-1})^2=0= > x^3+2x-7=2x-1= > x^3=6= > x=\sqrt[3]{6}= > y=\frac{\sqrt[3]{6}-1}{2}$
-Nếu $a^2+ab+b^2-3a-3b+4=0$
Ta có $0=a^2+ab+b^2-3(a+b)+4\geq \frac{3(a+b)^2}{4}-3(a+b)+4=\frac{3}{4}(a+b-2)^2+1\geq 1> 0= >$ vô lý
Vậy $(x,y)=(\sqrt[3]{6},\frac{\sqrt[3]{6}-1}{2})$