$\sqrt[]{5x²+10x+1}≥-x²-2x-7$
Điều kiện: $5x²+10x+1≥0$
⇔ \(\left[ \begin{array}{l}x≤\frac{-5-2.\sqrt[]{5}}{5}\\x≥\frac{-5+2.\sqrt[]{5}}{5}\end{array} \right.\)
Đặt $\sqrt[]{5x²+10x+1}=t$
⇒ $t²=5x²+10x+1$
⇔ $t²=5.( x²+2x+7)-34$
⇔ $-x²-2x-7=-\frac{t²+34}{5}$
⇒ $t≥-\frac{t²+34}{5}$
⇔ $t+0,2t²+6,8≥0$
Vì $Δ=1-4.0,2.6,8=-4,44<0$
⇒ $t+0,2t²+6,8>0$ $∀t$
⇒ $t∈R$
⇒ $x∈( -∞;\frac{-5-2.\sqrt[]{5}}{5}]U[\frac{-5+2.\sqrt[]{5}}{5};+∞)$
