Question

ai giúp mk trước 5h với ạ

Answer 1

Bài 2:

\(\sqrt{x^2+2x+4}=x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+2x+4=x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x=-2\end{matrix}\right.\)(vô nghiệm)

Vậy pt vô nghiệm

Answer 2

\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}=\dfrac{\sqrt{32}-\sqrt{12}}{-\left(\sqrt{48}-\sqrt{18}\right)}=-\dfrac{\sqrt{2}\left(\sqrt{16}-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{16}-\sqrt{6}\right)}=\dfrac{-\sqrt{2}}{\sqrt{3}}=\dfrac{-\sqrt{6}}{3}\)