Đáp án:
b, $\left \{ {{y=\frac{-1}{2}} \atop {x=\frac{3+\sqrt[]{2}}{2}}} \right.$
e, $\left \{ {{x=\frac{5-5\sqrt[]{3}-2\sqrt[]{5}}{3.\sqrt[]{5}.(1-\sqrt[]{3})}} \atop {y=\frac{\sqrt[]{5}+\sqrt[]{3}-1}{3}}} \right.$
Giải thích các bước giải:
a, $\left \{ {{(\sqrt[]{2}-1)x-y=\sqrt[]{2}} \atop {x+(\sqrt[]{2}+1)y=1}} \right.$
⇔ $\left \{ {{(\sqrt[]{2}-1).(\sqrt[]{2}+1)x-(\sqrt[]{2}+1)y=\sqrt[]{2}.(\sqrt[]{2}+1)} \atop {x+(\sqrt[]{2}+1)y=1}} \right.$
⇔ $\left \{ {{x-(\sqrt[]{2}+1)y=\sqrt[]{2}.(\sqrt[]{2}+1)} \atop {x+(\sqrt[]{2}+1)y=1}} \right.$
⇔ $\left \{ {{2.(\sqrt[]{2}+1)y=1-\sqrt[]{2}.(\sqrt[]{2}+1)} \atop {x=1-(\sqrt[]{2}+1)y}} \right.$
⇔ $\left \{ {{y=\frac{-1}{2}} \atop {x=\frac{3+\sqrt[]{2}}{2}}} \right.$
b, $\left \{ {{x\sqrt[]{5}-(1+\sqrt[]{3})y=1} \atop {(1-\sqrt[]{3})x+y\sqrt[]{5}=1}} \right.$
⇔ $\left \{ {{x\sqrt[]{5}.(1-\sqrt[]{3})-(1-\sqrt[]{3}).(1+\sqrt[]{3})y=1-\sqrt[]{3}} \atop {(1-\sqrt[]{3}).\sqrt[]{5}.x+y\sqrt[]{5}.\sqrt[]{5}=\sqrt[]{5}}} \right.$
⇔ $\left \{ {{x\sqrt[]{5}.(1-\sqrt[]{3})+2y=1-\sqrt[]{3}} \atop {(1-\sqrt[]{3}).\sqrt[]{5}.x+5y=\sqrt[]{5}}} \right.$
⇔ $\left \{ {{x\sqrt[]{5}.(1-\sqrt[]{3})+2y=1-\sqrt[]{3}} \atop {3y=\sqrt[]{5}+\sqrt[]{3}-1}} \right.$
⇔ $\left \{ {{x\sqrt[]{5}.(1-\sqrt[]{3})+2y=1-\sqrt[]{3}} \atop {y=\frac{\sqrt[]{5}+\sqrt[]{3}-1}{3}}} \right.$
⇔ $\left \{ {{x=\frac{5-5\sqrt[]{3}-2\sqrt[]{5}}{3.\sqrt[]{5}.(1-\sqrt[]{3})}} \atop {y=\frac{\sqrt[]{5}+\sqrt[]{3}-1}{3}}} \right.$
