\(\left(x-3\right)^2+2\left(x-1\right)\le x^2+3\)
\(\Leftrightarrow x^2-6x+9+2x-2\le x^2+3\)
\(\Leftrightarrow-4x\le-4\)
\(\Leftrightarrow x\ge1\)
Vậy \(S=\left\{x|x\ge1\right\}\)
Đúng 7
Bình luận (3)
\(\left(x-3\right)^2+2\left(x-1\right)\le x^2+3\)
\(\Leftrightarrow x^2-6x+9+2x-2-x^2-3\le0\)
\(\Leftrightarrow-4x+4\le0\)
\(\Leftrightarrow-4x\le-4\)
\(\Leftrightarrow x\ge1\)
Đúng 4
Bình luận (0)
