Ta có:
l\(4x^2+\left|3x+2\right|\)l \(=4x^2+2x+3\)
Vì \(\left\{{}\begin{matrix}4x^2\ge0\\\left|3x+2\right|\ge0\end{matrix}\right.\)\(\rightarrow\) l\(4x^2+\left|3x+2\right|\)l \(\ge0\)
\(\Rightarrow\) l\(4x^2+\left|3x+2\right|\)l \(=4x^2+\left|3x+2\right|\)
Khi đó \(PT\) trở thành:
\(4x^2+\left|3x+2\right|=4x^2+2x+3\)
\(\Leftrightarrow\left|3x+2\right|=2x+3\)\(\Leftrightarrow\left[{}\begin{matrix}3x+2=-2x-3\\3x+2=2x+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=-5\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy \(x=\pm1\)
Đúng 0
Bình luận (1)
