Có pt $(1)\Leftrightarrow (x-1)^3+3(x-1)=(y-2)^3+3(y-2)\Leftrightarrow x-1=y-2$$\Leftrightarrow x=y-1$ thay vào pt (2) đc
$(2) \Leftrightarrow y\sqrt{y+2}+(y+6)\sqrt{y+9}=y^2+4y-4$ (đk $y\geq-2$)
$\Leftrightarrow y(\sqrt{y+2}-3)+(y+6)(\sqrt{y+9}-4)=y^2-3y-28$
$\Leftrightarrow y.\frac{y-7}{\sqrt{y+2}+3}+(y+6).\frac{y-7}{\sqrt{y+9}+4}=(y-7)(y+4)$
$\Leftrightarrow \left[ {\begin{matrix} y=7(tm)\\ \frac{y}{\sqrt{y+2}+3}+\frac{y+6}{\sqrt{y+9}+4}=y+4 (3) \end{matrix}} \right.$
$+) y=7\Rightarrow x=6$
$+) $Ta cm đc pt (3) vô nghiệm
Thật vậy: ta có $\frac{y}{\sqrt{y+2}+3}\leq \frac{\left| {y} \right|}{\sqrt{y+2}+3}\leq \frac{\left| {y} \right|}{3}\Rightarrow ptvn$
cmtt: $\frac{y+6}{\sqrt{y+9}+4}\leq \frac{y+6}{4}< \frac{y+6}{3}$
$\Rightarrow VT(3)< \frac{\left| {y} \right|+y+6}{3}$
Lại có
-) Khi y>0 thì $VT(3)< \frac{\left| {y} \right|+y+6}{3}=\frac{2y+6}{3}<y+4=VP(3)\Rightarrow ptvn$
-) khi $y\leq 0$ thì $VT(3)< \frac{\left| {y} \right|+y+6}{3}=2\leq y+4=VP(3)(do y\geq -2)$