ĐK: x≥−52
BPT ⇔(x3+3x2+14x+15−2(x+2)√2x+5−3(x+2)√x2+5−3√5x2+7≤0
⇔x3+3x2−x−18−2(x+2)(√2x+5−3)−3(x+2)(√x2+5−3)+3−3√5x2+7≤0
⇔ Liên hợp ~~
⇔(x−2)[x2+5x+9−4(x+2)√2x+5+3−3(x+2)2√x2+5+3−5(x+2)9+33√5x2+7+(3√5x2+7)2≤0
Với x≥−52⇒{4(x+2)√2x+5+3<43(x+2)3(x+2)2√x2+5+3<35(x+2)5(x+2)..........<5(x+2)9⇒ f(x) >18x2+57x+12745>0
Do đó x−2≤0⇔x≤2
Kết hợp đc: x∈[−52;2]