a) Đặt 3√x+1=a;3√x−1=b=> PT ⇔a2+5b2=6ab⇔(a−b)(a−5b)=0TH1 a=b\Leftrightarrow x+1=x-1=> vô nghiệmTH2 a=5b⇔x+1=125(x−1)⇔x=6362
a) Đặt
3√x+1=a;3√x−1=b=> PT
⇔a2+5b2=6ab⇔(a−b)(a−5b)=0TH1
$a=b\Leftrightarrow x+1=x-1
$=> vô nghiệmTH2
a=5b⇔x+1=125(x−1)⇔x=6362