Hoặc
Ta có (4x – 1)10 = (4x – 1)8.
⇔ (4x – 1)10 – (4x – 1)8 = 0.
⇔ (4x – 1)8.[(4x – 1)2 – 1] = 0.
⇔4x−18=04x−12−1=0⇔4x−1=04x−1−14x−1+1=0⇔x=144x4x−2=0⇔x=14x=04x−2=0⇔x=14x=0x=12
Vậy x∈14;0;12