\(\left|x^2+2x+3\right|+\left|x-1\right|=6\)
\(\Leftrightarrow\left|x^2+2x+1+2\right|+\left|x-1\right|=6\)
\(\Leftrightarrow\left|\left(x+1\right)^2+2\right|+\left|x-1\right|=6\)
Do \(\left(x+1\right)^2+2\ge2\forall x\)
\(\Rightarrow\left|\left(x+1\right)^2+2\right|=\left(x+1\right)^2+2\)
Mà \(\left|\left(x+1\right)^2+2\right|+\left|x-1\right|=6\)
\(\Rightarrow\left(x+1\right)^2+2+\left|x-1\right|=6\)
\(\Rightarrow\left(x+1\right)^2+\left|x-1\right|=4\)
TH 1 : \(x\ge1\)
\(\Rightarrow x-1\ge0\)
\(\Rightarrow\left|x-1\right|=x-1\)
Mà \(\left(x+1\right)^2+\left|x-1\right|=4\)
\(\Rightarrow x^2+2x+1+x-1=4\)
\(\Rightarrow x^2+3x=4\)
\(\Rightarrow x^2+3x-4=0\)
\(\Rightarrow x^2+2x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{25}{4}=0\)
\(\Rightarrow\left(x+\dfrac{3}{2}\right)^2-\left(\dfrac{5}{2}\right)^2=0\)
\(\Rightarrow\left(x+\dfrac{3}{2}-\dfrac{5}{2}\right)\left(x+\dfrac{3}{2}+\dfrac{5}{2}\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-4\left(L\right)\end{matrix}\right.\left(1\right)\)
TH 2 : \(x< 1\)
\(\Rightarrow x-1< 0\)
\(\Rightarrow\left|x-1\right|=-x+1\)
Mà \(\left(x+1\right)^2+\left|x-1\right|=4\)
\(\Rightarrow x^2+2x+1-x+1=4\)
\(\Rightarrow x^2+x+2=4\)
\(\Rightarrow x^2+x-2=0\)
\(\Rightarrow x^2+2x\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{9}{4}=0\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2-\left(\dfrac{3}{2}\right)^2=0\)
\(\Rightarrow\left(x+\dfrac{1}{2}-\dfrac{3}{2}\right)\left(x+\dfrac{1}{2}+\dfrac{3}{2}\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\left(L,x< 1\right)\\x=-2\end{matrix}\right.\)(2)
Từ (1) (2)
\(\Rightarrow x\in\left\{1;-2\right\}\)
Vậy ...