Question

√x²+20x+25

Answer 1

ĐKXĐ: \(x^2+20x+25\ge0\)

=>\(x^2+20x+100-75\ge0\)

=>\(\left(x+10\right)^2\ge75\)

=>\(\left[\begin{array}{l}x+10\ge5\sqrt3\\ x+10\le-5\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge5\sqrt3-10\\ x\le-5\sqrt3-10\end{array}\right.\)

Ta có: \(\sqrt{x^2+20x+25}\ge0\forall x\) thỏa mãn ĐKXĐ

=>Dấu '=' xảy ra khi \(x^2+20x+25=0\)

=>\(x=-10\pm5\sqrt3\)