1.
a, -3x^2.(x^2 - 5x = 7)
= -3x^4 + 15 x^3 + 21x^2
b, (x-6).(-x-3)
= -x^2 - 3x + 6x + 18
= -x^2 - 3x +18
2
a, x^2 - 3x + xy - 3y
= (x^2 - 3x) + (xy - 3y)
= x.(x+3) + y.(x-3)
= (x+y).(x-3)
b, 16.(2x+3) - 9.(5x-2)^2
( câu này mik ko nghĩ ra. sorry )
3.
a, 2018x - 1 + 2019x.(1-2018x) = 0
-1.(1-2018x) + 2019x.(1-2018) = 0
(-1+2019x).(1-2018x) = 0
TH1: -1+2019x = 0
2019x = 1
x = 1/2019
TH2: 1 - 2018x = 0
2018x = 1
x = 1/2018
Vậy x=1/2018 hoặc x=1/2019
b, (x+2)^3 - x^2.(x-6) = 4
(x^3 + 3.x^2.2 + 3.x.2^2+2^3) + (x^3+6x^2) = 4
x^3 + 6x^2 + 12x + 18 + x^3 - 6x^2 = 4
(x^3-x^3) + (6x^2-6x^2) + 12x + 8 = 4
12x + 8 = 4
12x = 12
x = 1
Vậy x=1
