Question

Câu 7 ạ 🥲

Answer 1

Câu 7:

Theo đề, ta có:

\(\left\{{}\begin{matrix}a\sqrt{2}+b=2\\a+b=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\left(\sqrt{2}-1\right)=2-\sqrt{2}\\a+b=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\sqrt{2}\\b=0\end{matrix}\right.\)