Giải các hệ phương trình sau:

$a) \left\{\begin{array}{l}a\sqrt{5}-\left(1+\sqrt{3}\right)y=1\\ \left(1-\sqrt{3}\right)x+y\sqrt{5}=1\end{array}\right.; b) \left\{\begin{array}{l}\frac{2x}{x+1}+\frac{y}{y+1}=\sqrt{2}\\ \frac{x}{x+1}+\frac{3y}{y+1}=-1\end{array}\right.$

Giải: 

$$\begin{gathered} a) \left\{\begin{array}{l}a\sqrt{5}-\left(1+\sqrt{3}\right)y=1 \left(1\right)\\ \left(1-\sqrt{3}\right)x+y\sqrt{5}=1 \left(2\right)\end{array}\right. \\ \left(2\right)\iff y=\frac{1-\left(1-\sqrt{3}\right)x}{\sqrt{5}} \left(3\right) \end{gathered}$$

Thế (3) vào (1): 

$$\begin{gathered} x\sqrt{5}-\left(1+\sqrt{3}\right)\left[\frac{1-\left(1-\sqrt{3}\right)x}{\sqrt{5}}\right]=1 \\ \iff 5x-\left(1+\sqrt{3}\right)+\left(1-3\right)x=\sqrt{5} \\ \iff 3x=\sqrt{5}+\sqrt{3}+1 \\ \iff x=\frac{\sqrt{5}+\sqrt{3}+1}{3}\approx 1,66 \end{gathered}$$

Thay vào (3) được: $y=\frac{1-\left(1-\sqrt{3}\right){\displaystyle \frac{\sqrt{5}+\sqrt{3}+1}{3}}}{\sqrt{5}}\iff y=\frac{\sqrt{5}+\sqrt{3}-1}{3}$

Vậy hệ có nghiệm $\left(x;y\right)=\left(\frac{\sqrt{5}+\sqrt{3}+1}{3};\frac{\sqrt{5}+\sqrt{3}-1}{3}\right)$

b) Đặt $\frac{x}{x+1}=u$ và $\frac{y}{y+1}=v$, ta có hệ phương trình

$$\begin{gathered} \left\{\begin{array}{l}2u+v=\sqrt{2}\\ u+3v=-1\end{array}\right.\begin{array}{c}\left(1\right)\\ \left(2\right)\end{array} \\ \left(1\right)\iff v=\sqrt{2}-2u \left(3\right) \end{gathered}$$

Thay (3) vào (2): 

$$\begin{gathered} u+3\left(\sqrt{2}-2u\right)=-1 \\ \iff u+3\sqrt{2}-6u=-1\iff 5u=3\sqrt{2}+1 \\ \iff u=\frac{3\sqrt{2}+1}{5} \end{gathered}$$

Thay vào (3):

$v=\sqrt{2}-2.\frac{3\sqrt{2}+1}{5}\iff v=\frac{5\sqrt{2}-6\sqrt{2}-2}{5}\iff v=\frac{-2-\sqrt{2}}{5}$ nên hệ đã cho tương đương với:

$$\begin{gathered} \left\{\begin{array}{l}\frac{x}{x+1}=\frac{1+3\sqrt{2}}{5}\\ \frac{y}{y+1}=\frac{-2-\sqrt{2}}{5}\end{array}\right.\iff \left\{\begin{array}{l}5x=\left(1+3\sqrt{2}\right)\left(x+1\right)\\ 5y=\left(-2-\sqrt{2}\right)\left(y+1\right)\end{array}\right. \\ \iff \left\{\begin{array}{l}\left(4-3\sqrt{2}\right)x=1+3\sqrt{2}\\ \left(7+\sqrt{2}\right)y=-2-\sqrt{2}\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{1+3\sqrt{2}}{4-3\sqrt{2}}\\ y=-\frac{2+\sqrt{2}}{7+\sqrt{2}}\end{array}\right. \end{gathered}$$

Vậy hệ phương trình có nghiệm $\left(x;y\right)=\left(\frac{1+3\sqrt{2}}{4-3\sqrt{2}};-\frac{2+\sqrt{2}}{7+\sqrt{2}}\right)$