a, \(\left|x^2+2x+1\right|=\left|x^2-x+2\right|\)
\(\Leftrightarrow x^2+2x+1=\pm\left(x^2-x+2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=x^2-x+2\\x^2+2x+1=-x^2+x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1-x^2+x-2=0\\x^2+2x+1+x^2-x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\left(1\right)\\2x^2-x+3=0\left(2\right)\end{matrix}\right.\)
Giải (1) có : \(3x-1=0\) \(\Leftrightarrow x=\dfrac{1}{3}\)
Giải (2) có : \(2x^2-x+3=0\Rightarrow pt\) \(\text{vô nghiệm }\)
Vậy nghiệm của pt là \(\dfrac{1}{3}\)
b, \(\left|x^2+2x+3\right|=2x+3\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\ge0\text{(1) }\\\left[{}\begin{matrix}x^2+2x+3=2x+3\text{(2) }\\x^2+2x+3=-\left(2x+3\right)\text{(3) }\end{matrix}\right.\end{matrix}\right.\)
Giải (1) có : \(2x+3\ge0\Rightarrow x\ge\dfrac{3}{2}\)
Giải (2) ta có : \(x^2+2x+3=2x+3\)
\(\Leftrightarrow x^2=0\Rightarrow x=0\left(TM\right)\)
Giải (3) ta có : \(x^2+2x+3=-2x-3\)
\(\Leftrightarrow x^2+4x+6=0\Rightarrow pt\) vô nghiêm
Vậy pt có nghiệm là x = 0
a, \(\left|x^2+2x+1\right|=\left|x^2-x+2\right|\)
\(\Leftrightarrow\left|\left(x+1\right)^2\right|=\left|\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right|\)
\(\Leftrightarrow\left(x+1\right)^2=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x-\dfrac{1}{2}\right)^2=\dfrac{7}{4}\)
\(\Leftrightarrow\left(x+1+x-\dfrac{1}{2}\right)\left(x+1-x+\dfrac{1}{2}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\left(2x+\dfrac{1}{2}\right)\cdot\dfrac{3}{2}=\dfrac{7}{4}\)
\(\Leftrightarrow2x+\dfrac{1}{2}=\dfrac{7}{6}\Leftrightarrow x=\dfrac{\dfrac{7}{6}-\dfrac{1}{2}}{2}=\dfrac{1}{3}\)
Vậy......................