Giải thích các bước giải:
Ta có :
$(\sqrt{2}-1)x-y=\sqrt{2}$
$\to (\sqrt{2}+1)(\sqrt{2}-1)x-(\sqrt{2}-1)y=\sqrt{2}(\sqrt{2}-1)$
$\to (2-1)x-(\sqrt{2}-1)y=2-\sqrt{2}$
$\to x-(\sqrt{2}-1)y=2-\sqrt{2}$
$\to x=(\sqrt{2}-1)y+2-\sqrt{2}$
Mà $x+(\sqrt{2}+1)y=1$
$\to (\sqrt{2}-1)y+2-\sqrt{2}+(\sqrt{2}+1)y=1$
$\to (\sqrt{2}-1)y+(\sqrt{2}+1)y=\sqrt{2}-1$
$\to (\sqrt{2}-1+\sqrt{2}+1)y=\sqrt{2}-1$
$\to 2\sqrt{2}.y=\sqrt{2}-1$
$\to y=\dfrac{\sqrt{2}-1}{2\sqrt{2}}$
$\to x=1-(\sqrt{2}+1)y=1-(\sqrt{2}+1).\dfrac{\sqrt{2}-1}{2\sqrt{2}}1-\dfrac{2-1}{2\sqrt{2}}=1-\dfrac{1}{2\sqrt{2}}$
