Question

\(\int^{\sqrt{3}x-y=\sqrt{6}}_{\sqrt{2}x+y=2}\) giải hệ phương trình

Answer 1

Cộng 2 vế ta đc : \(\left(\sqrt{2}+\sqrt{3}\right)x=2+\sqrt{6}\Rightarrow x=\sqrt{2}\)

Thay x = \(\sqrt{2}\) vào \(\sqrt{2}\) x + y = 2 ta đc:

\(\sqrt{2}.\sqrt{2}+y=2\Rightarrow2+y=2\Rightarrow y=0\)

Vậy (x;y) = (\(\sqrt{2}\) ; 0)

 

Answer 2

\(\int^{\sqrt{3}x-y=\sqrt{6}}_{\left(\sqrt{3}x-y\right)+\left(\sqrt{2}x+y\right)=\sqrt{6}+2}\Leftrightarrow\int^{\sqrt{3}x-y=\sqrt{6}}_{\left(\sqrt{3}+\sqrt{2}\right)x=\sqrt{6}+2}\Leftrightarrow\int^{y=0}_{x=\sqrt{2}}\)

Vậy \(\left(x;y\right)=\left(\sqrt{2};0\right)\)