Question

giúp em bài 4 với ạ, em cảm ơn

Answer 1

\(x-\sqrt{2x+3}=0\)         \(ĐK:x\ge\dfrac{-3}{2}\)

\(\Leftrightarrow x=\sqrt{2x+3}\left(x\ge0\right)\)

\(\Leftrightarrow x^2=2x+3\)

\(\Leftrightarrow x^2-2x-3=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=-1\left(KTM\right)\end{matrix}\right.\)

Vậy.........