1. Giải các hệ phương trình sau:
a, $\left\{\begin{matrix}8y-x=4 & & \\ 2x-21y=2 & & \end{matrix}\right.$ b, $\left\{\begin{matrix}\frac{y}{4}-\frac{x}{5}=6 & & \\ \frac{x}{15}+\frac{y}{12}=0 & & \end{matrix}\right.$
c, $\left\{\begin{matrix}x\sqrt{2}+y=3+\sqrt{2} & & \\ -x+(\sqrt{2}-1)y=1-\sqrt{2} & & \end{matrix}\right.$ d, $\left\{\begin{matrix}x\sqrt{2}-y\sqrt{3}=5 & & \\ x+y=2\sqrt{2} & & \end{matrix}\right.$
Bài Làm:
a, $\left\{\begin{matrix}8y-x=4 & & \\ 2x-21y=2 & & \end{matrix}\right.$
Từ pt 8y - x = 4 => x = 8y - 4 (1)
Thay (1) vào phương trình 2x - 21y = 2 ta có:
2.(8y - 4)- 21y = 2 <=> -5y = 10 <=> y = -2
Thay y = -2 vào (1) ta có: x = 8.(-2) - 4 = -20
Vậy hệ có nghiệm (x; y) = (-20; -2)
b, $\left\{\begin{matrix}\frac{y}{4}-\frac{x}{5}=6 & & \\ \frac{x}{15}+\frac{y}{12}=0 & & \end{matrix}\right.$ <=> $\left\{\begin{matrix}5y-4x=120 & & \\ 4x+5y=0 & & \end{matrix}\right.$
<=> $\left\{\begin{matrix}5y-4x=120 & & \\ y=-\frac{4}{5}x & & \end{matrix}\right.$ <=> $\left\{\begin{matrix}5.\frac{-4}{5}x-4x=120 & & \\ y=\frac{-4}{5}x & & \end{matrix}\right.$ <=> $\left\{\begin{matrix}-8x=120 & & \\ y=\frac{-4}{5}x & & \end{matrix}\right.$
<=> $\left\{\begin{matrix}x=-15 & & \\ y=\frac{-4}{5}.(-15) & & \end{matrix}\right.$ <=> $\left\{\begin{matrix}x=-15 & & \\ y=12 & & \end{matrix}\right.$
Vậy hệ có nghiệm (x; y) = (-15; 12)
c, $\left\{\begin{matrix}x\sqrt{2}+y=3+\sqrt{2} & & \\ -x+(\sqrt{2}-1)y=1-\sqrt{2} & & \end{matrix}\right.$ <=> $\left\{\begin{matrix}x\sqrt{2}+y=3+\sqrt{2} & & \\ x=(\sqrt{2}-1)y-1+\sqrt{2} & & \end{matrix}\right.$
<=> $\left\{\begin{matrix}[(\sqrt{2}-1)y-1+\sqrt{2}].\sqrt{2}+y=3+\sqrt{2} & & \\ x=(\sqrt{2}-1)y-1+\sqrt{2} & & \end{matrix}\right.$
<=> $\left\{\begin{matrix}(3-\sqrt{2})y=1+2\sqrt{2} & & \\x=(\sqrt{2}-1)y-1+\sqrt{2} & & \end{matrix}\right.$ <=> $\left\{\begin{matrix}y=\frac{1+2\sqrt{2}}{3-\sqrt{2}} & & \\x=(\sqrt{2}-1)y-1+\sqrt{2} & & \end{matrix}\right.$
<=> $\left\{\begin{matrix}y=1+\sqrt{2} & & \\x=(\sqrt{2}-1).(1+\sqrt{2})-1+\sqrt{2} & & \end{matrix}\right.$ <=> $\left\{\begin{matrix}y=1+\sqrt{2} & & \\x=\sqrt{2} & & \end{matrix}\right.$
Vậy hệ phương trình có nghiệm (x; y) = ($\sqrt{2}; 1+\sqrt{2}$)
d, $\left\{\begin{matrix}x\sqrt{2}-y\sqrt{3}=5 & & \\ x+y=2\sqrt{2} & & \end{matrix}\right.$ <=> $\left\{\begin{matrix}x\sqrt{2}-y\sqrt{3}=5 & & \\ y=2\sqrt{2}-x & & \end{matrix}\right.$
<=> $\left\{\begin{matrix}x\sqrt{2}-(2\sqrt{2}-x)\sqrt{3}=5 & & \\ y=2\sqrt{2}-x & & \end{matrix}\right.$ <=> $\left\{\begin{matrix}x(\sqrt{2}+\sqrt{3})=5+2\sqrt{6} & & \\ y=2\sqrt{2}-x & & \end{matrix}\right.$
<=> $\left\{\begin{matrix}x=\frac{5+2\sqrt{6}}{\sqrt{2}+\sqrt{3}} & & \\ y=2\sqrt{2}-x & & \end{matrix}\right.$ <=> $\left\{\begin{matrix}x=\sqrt{2}+\sqrt{3} & & \\ y=2\sqrt{2}-x & & \end{matrix}\right.$ <=> $\left\{\begin{matrix}x=\sqrt{2}+\sqrt{3} & & \\ y=2\sqrt{2}-(\sqrt{2}+\sqrt{3}) & & \end{matrix}\right.$
<=> $\left\{\begin{matrix}x=\sqrt{2}+\sqrt{3} & & \\ y=\sqrt{2}-\sqrt{3} & & \end{matrix}\right.$
Vậy hệ phương trình có nghiệm (x; y) = ($\sqrt{2}+\sqrt{3}; \sqrt{2}-\sqrt{3}$)