Ta có: \(\left|x-2\right|+\left|5+x^2\right|=7\)
\(\Leftrightarrow\left|x-2\right|+x^2+5-7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2+x^2+5-7=0\left(x\ge2\right)\\2-x+x^2+5-7=0\left(x< 2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-4=0\\x^2-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\frac{1}{2}\right)^2=\frac{17}{4}\\x\left(x-1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{\sqrt{17}}{2}\\x+\frac{1}{2}=-\frac{\sqrt{17}}{2}\\x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{17}-1}{2}\left(loại\right)\\x=-\frac{\sqrt{17}+1}{2}\left(loại\right)\\x=0\left(nhận\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={0;1}
Vì 5+x2>0 \(\forall x\in R\) nên \(\left|5+x^2\right|=5+x^2\)
Có: \(\left|x-2\right|=\)\(\left\{{}\begin{matrix}x-2,\:khi\:x>2\\-x+2,\:khi\:x< 2\end{matrix}\right.\)
TH1: x>2
Có: x-2+5+x2=7
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1+\sqrt{17}}{2}\\x=\frac{1-\sqrt{17}}{2}\end{matrix}\right.\) (KTM vì nhỏ hơn 2)
TH2: x<2
Có: -x+2+5+x2=7
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)(TM)
Vậy nghiệm của pt là x=1,x=0
