b) 3(x - 2) = x(x - 2)
3(x - 2) - x(x - 2) = 0
=> (x - 2)(3 - x) = 0
TH1 : x - 2 = 0
=> x = 2
TH2 : 3 - x = 0
=> x = 3
Vậy x = 2 hoặc 3
c) (x + 3)7 = 4(x + 3) . 5
=> (x + 3)7 = 20(x + 3)
=> (x + 3)7 - 20(x + 3) = 0
\(\Rightarrow\left(x+3\right)\left[\left(x+3\right)^6-20\right]=0\)
TH1 : x + 3 = 0
=> x = -3
TH2 : (x + 3)6 - 20 = 0
=> (x + 3)6 = 20 (loại)
Vậy x = -3
d) x - 3.(x - 3) = x - 3
=> x - 3x + 9 = x - 3
=> x - 3x - x = -3 - 9
=> -3x = -12
=> x = 4
Vậy x = 4