Question

Giải phương trình

Answer 1

\(\sqrt{\left(3-x\right)^2}=2x+1\)

\(\left[{}\begin{matrix}3-x=2x+1\\3-x=-2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3-x-2x-1=0\\3-x+2x+1-0\end{matrix}\right.\)

\(\left[{}\begin{matrix}2-3x=0\\4+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)

Answer 2

\(\sqrt{9-6x+x^2}=2x+1\\ < =>\sqrt{\left(3-x\right)^2}-2x-1=0\\ < =>\left(3-x\right)-2x-1=0\\ < =>-3x+2=0\\ < =>x=\dfrac{2}{3}\)