tui lm tắt thui nhak!
Nhân 2 xế vs √x+1>0Đặt √x+1=a
BPT tg đg: 2x2(x2+1)+(x+2)a2≥a5x+4ax
⇔2x2(x+1)2−4x3+(x+2)a2≥a5x+4ax
⇔2a4x2−4x3+(x+2)a2−a5x−4ax≥0
⇔a4x(2x−a)+x(a2−4x2)+2a(a−2x)≥0
⇔(a−2x)(−a4x+a3+a+2x2)≥0
⇔(a−2x)(a3+a−x3−x)≥0
⇔(a−2x)(a−x)(a2+ax+x2+1)≥0
~~ dễ rồi...