Đáp án:

$\left(x;y\right)=\left(3;4\right),\left(7;8\right)$

...

$\left(x;y\right)=\left(5;4\right)$

Giải thích các bước giải:

$\begin{cases}x^3-y^3+3x^2+6x-3y+4=0\,\,\,\left(1\right)\\2\sqrt{x+y-6}+2\sqrt{3x-5}=y^2-2y-8x+22\,\,\,\left(2\right)\end{cases}$

Điều kiện: $x\ge \dfrac{5}{3}\,\,;\,\,x+y\ge 6$

$\left( 1 \right)\Leftrightarrow {{x}^{3}}+3{{x}^{2}}+6x+4={{y}^{3}}+3y$

$\Leftrightarrow \left( {{x}^{3}}+3{{x}^{2}}+3x+1 \right)+\left( 3x+3 \right)={{y}^{3}}+3y$

$\Leftrightarrow {{\left( x+1 \right)}^{3}}+3\left( x+1 \right)={{y}^{3}}+3y$

$\Leftrightarrow {{t}^{3}}+3t={{y}^{3}}+3y$   với   $t=x+1$

$\Leftrightarrow \left( t-y \right)\left( {{t}^{2}}+ty+{{y}^{2}} \right)+3\left( t-y \right)=0$

$\Leftrightarrow \left( t-y \right)\left( {{t}^{2}}+ty+{{y}^{2}}+3 \right)=0$

$\Leftrightarrow t-y=0$   (vì ${{t}^{2}}+ty+{{y}^{2}}+3>0$)

$\Leftrightarrow t=y$

$\Leftrightarrow x+1=y$

Thay $y=x+1$ vào PT $\left( 2 \right)$

$\left( 2 \right)\Leftrightarrow 2\sqrt{2x-5}+2\sqrt{3x-5}={{x}^{2}}-8x+21$

$\Leftrightarrow {{x}^{2}}-8x+21-2\sqrt{2x-5}-2\sqrt{3x-5}=0$

$\Leftrightarrow \left( {{x}^{2}}-10x+21 \right)+\left[ \left( x-1 \right)-2\sqrt{2x-5} \right]+\left[ \left( x+1 \right)-2\sqrt{3x-5} \right]=0$

$\Leftrightarrow \left( {{x}^{2}}-10x+21 \right)+\dfrac{{{\left( x-1 \right)}^{2}}-4\left( 2x-5 \right)}{x-1+2\sqrt{x-5}}+\dfrac{{{\left( x+1 \right)}^{2}}-4\left( 3x-5 \right)}{x+1+2\sqrt{3x-5}}=0$

$\Leftrightarrow \left( {{x}^{2}}-10x+21 \right)+\dfrac{{{x}^{2}}-10x+21}{x-1+2\sqrt{x-5}}+\dfrac{{{x}^{2}}-10x+21}{x+1+2\sqrt{3x-5}}=0$

$\Leftrightarrow \left( {{x}^{2}}-10x+21 \right)\left( 1+\dfrac{1}{x-1+2\sqrt{x-5}}+\dfrac{1}{x+1+2\sqrt{3x-5}} \right)=0$

$\Leftrightarrow {{x}^{2}}-10x+21=0$   (vế sau luôn dương)

$\Leftrightarrow x=3$(nhận)   hoặc   $x=7$ (nhận)

Với $x=3\Rightarrow y=4$

Với $x=7\Rightarrow y=8$

Vậy $\left( x;y \right)=\left( 3;4 \right),\left( 7;8 \right)$

$\begin{cases}x+4+\sqrt{x^2+8x+17}=y+\sqrt{y^2+1}\,\,\,\left(1\right)\\x+\sqrt{y}+\sqrt{y+21}+1=2\sqrt{2y-3x}\,\,\,\left(2\right)\end{cases}$

Điều kiện: $y\ge 0\,\,;\,\,2y\ge 3x$

$\left( 1 \right)\Leftrightarrow \left( x+4-y \right)+\left( \sqrt{{{x}^{2}}+8x+17}-\sqrt{{{y}^{2}}+1} \right)=0$

$\Leftrightarrow \left( x+4-y \right)+\dfrac{\left( {{x}^{2}}+8x+17 \right)-\left( {{y}^{2}}+1 \right)}{\sqrt{{{x}^{2}}+8x+17}+\sqrt{{{y}^{2}}+1}}=0$

$\Leftrightarrow \left( x+4-y \right)+\dfrac{\left( {{x}^{2}}+8x+16 \right)-{{y}^{2}}}{\sqrt{{{x}^{2}}+8x+17}+\sqrt{{{y}^{2}}+1}}=0$

$\Leftrightarrow \left( x+4-y \right)+\dfrac{{{\left( x+4 \right)}^{2}}-{{y}^{2}}}{\sqrt{{{x}^{2}}+8x+17}+\sqrt{{{y}^{2}}+1}}=0$

$\Leftrightarrow \left( x+4-y \right)+\dfrac{\left( x+4-y \right)\left( x+4+y \right)}{\sqrt{{{x}^{2}}+8x+17}+\sqrt{{{y}^{2}}+1}}=0$

$\Leftrightarrow \left( x+4-y \right)\left( 1+\dfrac{x+4+y}{\sqrt{{{x}^{2}}+8x+17}+\sqrt{{{y}^{2}}+1}} \right)=0$

$\Leftrightarrow x+4-y=0$   (vế sau luôn dương)

$\Leftrightarrow y=x+4$

Thay $y=x+4$ vào PT $\left( 2 \right)$

$\left( 2 \right)\Leftrightarrow x+\sqrt{x+4}+\sqrt{x+25}+1=2\sqrt{8-x}$

PT có nghiệm xấu vô tỷ, chưa tìm ra lời giải

Nếu thay đổi đề chỗ $2\sqrt{2y-3x}$ thành $2\sqrt{4y-3x}$

Thì khi đó PT sẽ có nghiệm đẹp và giải được

$\begin{cases}3\left(x+2\right)\sqrt{x-y}=y\sqrt{x-y}+3x-y+6\,\,\,\left(1\right)\\\sqrt{3x+1}-\sqrt{5-y}=-x^3+5x^2+3\,\,\,\left(2\right)\end{cases}$

Điều kiện: $x\ge -\dfrac{1}{3}\,\,;\,\,y\le 5\,\,;\,\,x\ge y$

$\left( 1 \right)\Leftrightarrow 3\left( x+2 \right)\sqrt{x-y}-3\left( x+2 \right)=y\sqrt{x-y}-y$

$\Leftrightarrow 3\left( x+2 \right)\left( \sqrt{x-y}-1 \right)=y\left( \sqrt{x-y}-1 \right)$

$\Leftrightarrow 3\left( x+2 \right)=y$   hoặc   $\sqrt{x-y}=1$

$\Leftrightarrow y=3x+6$   hoặc   $y=x-1$

Thay $y=3x+6$ vào PT $\left( 2 \right)$

$\left( 2 \right)\Leftrightarrow \sqrt{3x+1}-\sqrt{-3x-1}=-{{x}^{3}}+5{{x}^{2}}+3$

Điều kiện $\begin{cases}3x+1\ge 0\\-3x-1\ge 0\end{cases}\Leftrightarrow\begin{cases}x\ge -\dfrac{1}{3}\\x\le -\dfrac{1}{3}\end{cases}\Leftrightarrow x=-\dfrac{1}{3}$

Thay $x=-\dfrac{1}{3}$ trở lại, ta được $0=\dfrac{97}{27}$ (vô lý)

Vậy phương trình vô nghiệm

Thay $y=x-1$ vào PT $\left( 2 \right)$

$\left( 2 \right)\Leftrightarrow \sqrt{3x+1}-\sqrt{6-x}=-{{x}^{3}}+5{{x}^{2}}+3$   (ĐK: $-\dfrac{1}{3}\le x\le 6$)

$\Leftrightarrow \left( \sqrt{3x+1}-4 \right)+\left( 1-\sqrt{6-x} \right)+\left( {{x}^{3}}-5{{x}^{2}} \right)=0$

$\Leftrightarrow \dfrac{\left( 3x+1 \right)-16}{\sqrt{3x+1}+4}+\dfrac{1-\left( 6-x \right)}{1+\sqrt{6-x}}+{{x}^{2}}\left( x-5 \right)=0$

$\Leftrightarrow \dfrac{3\left( x-5 \right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{1+\sqrt{6-x}}+{{x}^{2}}\left( x-5 \right)=0$

$\Leftrightarrow \left( x-5 \right)\left( \dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+{{x}^{2}} \right)=0$

$\Leftrightarrow x-5=0$   (Vế sau luôn dương)

$\Leftrightarrow x=5$

Với $x=5\Rightarrow y=4$

Vậy $\left( x;y \right)=\left( 5;4 \right)$