Giải các hệ phương trình:

$$\begin{gathered} a)\left\{\begin{array}{l}2\sqrt{x-1}-\sqrt{y-1}=1\\ \sqrt{x-1}+\sqrt{y-1}=2\end{array}\right.; \\ b) \left\{\begin{array}{l}{\left(x-1\right)}^2-2y=2\\ 3{\left(x-1\right)}^2+3y=1\end{array}\right. \end{gathered}$$

Giải:

a) Đặt $\sqrt{x-1}=X\ge 0, \sqrt{y-1}=Y\ge 0$

Ta được: $\left\{\begin{array}{l}2X-Y=1\\ X+Y=2\end{array}\right.\iff \left\{\begin{array}{l}3X=3\\ X+Y=2\end{array}\right.\iff \left\{\begin{array}{l}X=1\\ Y=1\end{array}\right.$

hay $\left\{\begin{array}{l}\sqrt{x-1}=1\\ \sqrt{y-1}=1\end{array}\right.\iff \left\{\begin{array}{l}x-1=1\\ y-1=1\end{array}\right.\iff \left\{\begin{array}{l}x=2\\ y=2\end{array}\right.$

b) Đặt ${\left(x-1\right)}^2=X\ge 0$

Ta được: 

$$\begin{gathered} \left\{\begin{array}{l}X-2y=2\\ 3X+3y=1\end{array}\right.\iff \left\{\begin{array}{l}3X-6y=6\\ 3X+3y=1\end{array}\right.\iff \left\{\begin{array}{l}3X-6y=6\\ -9y=5\end{array}\right. \\ \iff \left\{\begin{array}{l}3X=6+6\left(-\frac{5}{9}\right)\\ y=-\frac{5}{9}\end{array}\right.\iff \left\{\begin{array}{l}X=\frac{8}{9}\\ y=-\frac{5}{9}\end{array}\right. \end{gathered}$$

Từ $X=\frac{8}{9}$ ta được ${\left(x-1\right)}^2=\frac{8}{9}\iff x-1=\pm \frac{2\sqrt{2}}{3}\iff x=1\pm \frac{2\sqrt{2}}{3}$

Vậy hệ phương trình có nghiệm: $$\begin{gathered} x_1=1+\frac{2\sqrt{2}}{3}; y=-\frac{5}{9} \\ {x}_2=1-\frac{2\sqrt{2}}{3}; y=-\frac{5}{9} \end{gathered}$$