Question

Phân tích đa thức thành nhân tử :

A , \(X^3-4.X^2+4.X+4.X-8\)

B , \(4.X^2-25-\left(2.X-5\right).\left(2.X-7\right)\)

C , \(X^3+27+\left(X+3.\right)\left(X-9\right)\)

Giúp mình với

Answer 1

x^3 - 4x^2 + 4x + 4x - 8

= (X^3 - 8) - (4x^2 - 4x - 4x)

= (x - 2)(x^2 + 2x + 4) - 4x( x - 2)

= (x - 2)(x^2 + 2x + 4 - 4x)

= (x - 2)(x^2 - 2x + 4)

b) 4x^2 - 25 - (2x - 5)(2x-  7)

= (2x - 5)(2x + 5) - (2x - 5)(2x - 7)

= (2x - 5)(2x + 5 - 2x + 7)

= 12(2x - 5)

c) x^3 + 27 + (x + 3)(x - 9)

= (x+3)(x^2-3x+9) + (x + 3)(x - 9)

= (x + 3) (x ^2 -3x + 9 + x - 9)

= (x + 3)(x^2 - 2x) = x(x - 2)(x + 3)

Answer 2

dễ mà bạn ơi

Answer 3

Bạn giúp mình trả lời đi

Answer 4

a, x³-4x²+4x+4x-8

=(x³-8)-(4x²-4x-4x)

=(x³-2³)-(4x²-8x)

=(x-2)(x²+2x+4)-4x(x-2)

=(x-2)(x²+2x+4-4x)

=(x-2)(x²-2x+4)

b, 4x²-25-(2x-5)(2x-7)

=[(2x)²-(5)²]-(2x-5)(2x-7)

=(2x-5)(2x+5)-(2x-5)(2x-7)

=(2x-5)(2x+5-2x+7)

=(2x-5)12

=12(2x-5) 

c,x³+27+(x+3)(x-9)

=(x+3)(x²+3x+9)+(x+3)(x-9)

=(x+3)(x²-3x+9+x-9)

=(x+3)(x²-2x)

=x(x+3)(x-2)