$$\left \lbrace \begin{array}{l} \sqrt{x-2}-\sqrt{y-1}=1 \\ x+\sqrt{8x+y^2}=8 \end{array} \right.$$

Đề bài:
Giải hệ phương trình:
$$\left \lbrace \begin{array}{l} \sqrt{x-2}-\sqrt{y-1}=1 \\ x+\sqrt{8x+y^2}=8 \end{array} \right.$$

Lời giải:
Đánh số hai phương trình của hệ:
\begin{align} \label{eq:139.1} \sqrt{x-2}-\sqrt{y-1}=1 \\ \label{eq:139.2} x+\sqrt{8x+y^2}=8 \end{align}

Điều kiện: $x \geqslant 2$; $y \geqslant 1$
\begin{align} \eqref{eq:139.1} & \Leftrightarrow 1=\sqrt{x-2}-\sqrt{y-1}\leqslant \sqrt{x-2} \quad \left(\sqrt{y-1}\geqslant 0\right) \nonumber \\ & \Rightarrow 1\leqslant x-2 \text{ (hai vế không âm)} \nonumber \\ & \Leftrightarrow x\geqslant 3 \nonumber \end{align}

Ta có: $x+\sqrt{8x+y^2} \geqslant 3+\sqrt{2.3+1^2} =8 \quad \left(x\geqslant 3; y\geqslant 1\right)$
\begin{equation} \label{eq:139.3} x+\sqrt{8x+y^2}\geqslant 8 \end{equation}
$$\eqref{eq:139.2}, \eqref{eq:139.3} \Rightarrow \left\lbrace \begin{array}{l} x=3 \\ y=1 \end{array} \right.$$
$$S=\left\lbrace \left( 3; 1 \right) \right\rbrace$$

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Ta vẫn còn một cách giải khác như sau:

Điều kiện $x \geqslant 2 \wedge y \geqslant 1$

\begin{align} \eqref{eq:139.1} & \Leftrightarrow \sqrt{x-2} =1+\sqrt{y-1} \nonumber \\ & \Leftrightarrow x-2 =1+2\sqrt{y-1}+y-1 \text{ (2 vế không âm)} \nonumber \\ & \Leftrightarrow x-y-2 =2\sqrt{y-1} \nonumber \\ & \Rightarrow x^2+y^2+4-2xy-4x+4y =4y-4 \quad \left(x-2 \geqslant y\right) \nonumber \\ & \Leftrightarrow x^2+y^2-2xy-4x+8=0 \nonumber \\ & \Leftrightarrow x^2-2\left(y+2\right)x+y^2+8=0 \nonumber \\ \Delta'_x &=(y+2)^2-\left(y^2+8\right) \nonumber \\ &=y^2+4y+4-y^2-8 \nonumber \\ &=4y-4 \nonumber \\ \Rightarrow \sqrt{\Delta'_x} & =2\sqrt{y-1} \end{align}

Ta có hệ hai phương trình hệ quả sau:
\begin{align} \label{eq:139.4} x=y+2+2\sqrt{y-1} \\ \label{eq:139.5} x=y+2-2\sqrt{y-1} \end{align}

Thay \eqref{eq:139.4} vào \eqref{eq:139.1} ta được:
\begin{align} & \phantom{\Leftrightarrow} -y-2\sqrt{y-1}-2+8=\sqrt{8\left(y+2\sqrt{y-1} +2\right)+y^2} \nonumber \\ & \Leftrightarrow -y-2\sqrt{y-1}+6=\sqrt{8\left(y+2\sqrt{y-1} +2\right)+y^2} \nonumber \\ & \Rightarrow y^2+4y-4+36+4y\sqrt{y-1}-24\sqrt{y-1}-12y=8\left(y+2\sqrt{y-1} +2\right)+y^2 \left(-y-2\sqrt{y-1}+6 \geqslant 0\right)\nonumber \\ & \Leftrightarrow 4y-4+36+4y\sqrt{y-1}-24\sqrt{y-1}-12y-8y-16\sqrt{y-1} -16=0 \nonumber \\ & \Leftrightarrow 4y\sqrt{y-1}-40\sqrt{y-1}-16y+16=0 \nonumber \end{align}

Đặt $\sqrt{y-1}=a \quad (a\geqslant 0) \Rightarrow y=a^2+1$
\begin{align} & \Rightarrow 4a(a^2+1)-40a-16(a^2+1)+16 =0 \nonumber \\ & \Leftrightarrow 4a^3+4a-40a-16a^2-16+16 =0 \nonumber \\ & \Leftrightarrow 4a^3-16a^2-36a=0 \nonumber \\ & \Leftrightarrow a\left(4a^2-16a-36\right)=0 \nonumber \\ & \Leftrightarrow \left[ \begin{array}{l} a=0 \\ a^2-4a-9=0 \quad (\Delta'=4+9=13>0) \end{array} \right. \nonumber \\ & \Leftrightarrow \left[ \begin{array}{l} a=0 \\ a=2+\sqrt{13} \\ a=2-\sqrt{13}<0 \end{array} \right. \nonumber \end{align}

Thay \eqref{eq:139.5} vào \eqref{eq:139.1} ta được:
\begin{align} & \phantom{\Leftrightarrow} -y+2\sqrt{y-1}-2+8=\sqrt{8\left(y-2\sqrt{y-1} +2\right)+y^2} \nonumber \\ & \Leftrightarrow -y+2\sqrt{y-1}+6=\sqrt{8\left(y-2\sqrt{y-1} +2\right)+y^2} \nonumber \\ & \Rightarrow y^2+4y-4+36-4y\sqrt{y-1}+24\sqrt{y-1}-12y=8\left(y-2\sqrt{y-1} +2\right)+y^2 \left(-y+2\sqrt{y-1}+6 \geqslant 0\right)\nonumber \\ & \Leftrightarrow 4y-4+36-4y\sqrt{y-1}+24\sqrt{y-1}-12y-8y+16\sqrt{y-1} -16=0 \nonumber \\ & \Leftrightarrow 4y\sqrt{y-1}-40\sqrt{y-1}+16y-16=0 \nonumber \\ & \Rightarrow 4a(a^2+1)-40a+16(a^2+1)-16 =0 \text{ (với $\sqrt{y-1}=a \quad (a\geqslant 0) \Rightarrow y=a^2+1$)} \nonumber \\ & \Leftrightarrow 4a^3+4a-40a+16a^2+16-16 =0 \nonumber \\ & \Leftrightarrow 4a^3+16a^2-36a=0 \nonumber \\ & \Leftrightarrow a\left(4a^2+16a-36\right)=0 \nonumber \\ & \Leftrightarrow \left[ \begin{array}{l} a=0 \\ a^2+4a-9=0 \quad (\Delta'=4+9=13>0) \end{array} \right. \nonumber \\ & \Leftrightarrow \left[ \begin{array}{l} a=0 \\ a=-2+\sqrt{13} \\ a=-2-\sqrt{13}<0 \end{array} \right. \nonumber \end{align}

Với $a=0 \Rightarrow y=1 \stackrel{\eqref{eq:139.1}}{\Rightarrow} \sqrt{x-2}=1 \Leftrightarrow x=3$

Với $a=2+\sqrt{13} \Rightarrow y=18+4\sqrt{13} \stackrel{\eqref{eq:139.1}}{\Rightarrow} \sqrt{x-2}=1+\sqrt{17+4\sqrt{13}}$
$$\Leftrightarrow x=2+\sqrt{17+4\sqrt{13}}+20+4\sqrt{13} \Leftrightarrow x=24+6\sqrt{13} \text{ (không thỏa mãn $-y-2\sqrt{y-1}+6 \geqslant 0$)}$$

Với $a=-2+\sqrt{13} \Rightarrow y=18-4\sqrt{13} \stackrel{\eqref{eq:139.1}}{\Rightarrow} \sqrt{x-2}=1+\sqrt{17-4\sqrt{13}}$
$$\Leftrightarrow x=2+\sqrt{17-4\sqrt{13}}+20-4\sqrt{13} \Leftrightarrow x=16-2\sqrt{13} \text{ (không thỏa mãn $-y+2\sqrt{y-1}+6 \geqslant 0$)}$$
$$S=\left\lbrace \left( 3; 1 \right) \right\rbrace$$