Ta có pt ⇔3x2+3x+5.(x+1−3√x3+1)=0 pt ⇔3x2+3x+5.(x+1)3−(x3+1)(x+1)2+(x+1).3√x3+1=0 pt ⇔x.(x+1).(3+5.3(x+1)2+(x+1).3√x3+1+(3√x3+1)2)=0Dễ thấy 3+5.3(x+1)2+(x+1).3√x3+1+(3√x3+1)2>0∀x∈R pt x(x+1)=0 nên nghiệm pt là x=0; x=-1
Ta có pt ⇔3x2+3x+5.(x+1−3√x3+1)=0 pt ⇔3x2+3x+5.(x+1)3−(x3+1)(x+1)2+(x+1).3√x3+1=0 pt ⇔x.(x+1).(3+5.3(x+1)2+(x+1).3√x3+1+(3√x3+1)2)=0Dễ thấy 3+5.3(x+1)2+(x+1).3√x3+1+(3√x3+1)2>0∀x∈R pt $ \Leftrightarrow x(x+1)=0$ nên nghiệm pt là x=0; x=-1