Question

Câu 2 ghi Kết quả thôi khỏi trình bày bước giải

Answer 1

3:

Tọa độ giao là:

\(\left\{{}\begin{matrix}x\left(\sqrt{2}+1-1+\sqrt{2}\right)=\sqrt{7}-\sqrt{5}\\y=\left(\sqrt{2}+1\right)x+\sqrt{5}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{\sqrt{7}-\sqrt{5}}{2\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{10}}{4}\\y=\left(\sqrt{2}+1\right)\cdot\dfrac{\sqrt{14}-\sqrt{10}}{4}+\sqrt{5}=\dfrac{2\sqrt{7}+2\sqrt{5}+\sqrt{14}-\sqrt{10}}{4}\end{matrix}\right.\)

=>\(m=y+\left(\sqrt{3}+\sqrt{5}\right)x\)

\(=\dfrac{2\sqrt{7}+2\sqrt{5}+\sqrt{14}-\sqrt{10}+\sqrt{42}-\sqrt{30}+\sqrt{70}-5\sqrt{2}}{4}\)