(x-2)^8=(x-2)^6
=>(x-2)^8-(x-2)^6=0
=>(x-2)^6.(x-2)^2-(x-2)^6=0
=>(x-2)^6 . [(x-2)^2-1]=0
=>(x-2)^6=0=>x-2=0=>x=2
hoặc (x-2)^2-1=0=>(x-2)^2=1=>x-2=1=>x=3
vậy x E {2;3}
(x - 2)8 = (x - 2)6
=> (x - 2)8 - (x - 2)6 = 0
(x - 2)6 . (x - 2)2 - (x - 2)6 . 1 = 0
(x - 2)6 . [(x - 2)2 - 1] = 0
=> (x - 2)6 = 0 hoặc (x - 2)2 - 1 = 0
1, (x - 2)6 = 0 2, (x - 2)2 - 1 = 0
(x - 2)6 = 06 (x - 2)2 = 1
x - 2 = 0 (x - 2)2 = 12
x = 2 x - 2 = 1
x = 3
Vậy x \(\in\) {2; 3}
x=2;3
Tik cho mk nha.............cảm ơn rất nhiều
