a. \(\left|x^2+4x-5\right|=x+2\left(ĐK:x\ge-2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+4x-5=x+2\\x^2+4x-5=-x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x-7=0\\x^2+5x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{37}}{2}\\x=\dfrac{-3-\sqrt{37}}{2}\left(l\right)\\x=\dfrac{-5+\sqrt{37}}{2}\\x=\dfrac{-5-\sqrt{37}}{2}\left(l\right)\end{matrix}\right.\)
Vậy........
b. \(\left|2x-4\right|-x^2+x-6=0\)
\(\Leftrightarrow\left|2x-4\right|=x^2-x+6\left(ĐK:x\in R\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-4=x^2-x+6\\2x-4=-x^2+x-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x^2+3x-10=0\\x^2-x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-\left(x-\dfrac{3}{2}\right)^2-\dfrac{31}{4}=0\left(l\right)\\\left(x+\dfrac{1}{2}\right)+\dfrac{7}{4}=0\left(l\right)\end{matrix}\right.\)
Vậy pt vô nghiệm
