(x2-11x+28)(x2-7x+10)-72=0
(x-2)(x-7)(x-4)(x-5)-72=0
(x2-9x+14)(x2-9x+20)-72=0
Đặt x2-9x+14=a, ta có
a(a+6)-72=0
a2+6a-72=0
a2+6a+9-81=0
(a+3)2=81
=> a+3=9 => a=6=> x2-9x+14=6
=>x2-9x+8=0
=> (x-1)(x-8)=0
=> x=1 hoặc x=8
Vậy............ !!!!!