Question

Hộ mình nhé =))

Phân tích đa thức thành nhân tử

a,  x³-x²-4x²+8x-4                                        e,  4x²+4x-3                                      

b,  4x²-25-(2x-5)(2x+7)                                 f,  4x²+4x²y²-8y⁴

c,  x²-4xy+4y²-9z²                                             g, (x²+x+1)(x²+x+2)-12

d, x2-7x+6                                                     h, 4x²y²-(x2+y²-z²)²

                                                                       i,x³-8-(x-2)(x+4)

***Các bạn làm đc đến đâu thì làm nhé, mình cảm ơn

Answer 1

Đăng từng câu  thôi 

Answer 2

Answer 3

a,  x³-x²-4x²+8x-4    

=x³-4x²+4x-x²+4x-4

=x(x²-4x+4)-(x²-4x+4)

=(x²-4x+4)(x-1)

=(x-2)²(x-1)

b,  4x²-25-(2x-5)(2x+7)

=4x²-5²-(2x-5)(2x+7)

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

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

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

=-2(2x-5)

c,  x²-4xy+4y²-9z²  

=(x²+4xy+4y²)-9z²

=(x+2y)²-(3z²)

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

d, x²-7x+6      

=x²-6x-x+6

=x(x-6)-(x-6)

=(x-6)(x-1)

e,  4x²+4x-3

=4x²+6x-2x-3

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

=(2x-1)(2x+3)

f,  4x²+4x²y²-8y⁴

=4(x²+x²y²-2y⁴)

g, (x²+x+1)(x²+x+2)-12

Đặt t=x²+x+1 ta có:

t(t+1)-12=t²+t-12

=t²-3t+4t-12

=t(t-3)+4(t-3)

=(t+4)(t-3)

=(x²+x+1+4)(x²+x+1-3)

=(x²+x+5)(x²+x-2)

=(x²+x+5)[x²+2x-x-2]

=(x²+x+5)[x(x+2)-(x+2)]

=(x²+x+5)(x-1)(x+2)

h, 4x²y²-(x²+y²-z²)²

=(2xy)²-(x²+y²-z²)²

=(2xy+x²+y²-z²)(z²-y²-x²+2xy)

=(xz+yz-z²+xy+y²-yz+x²+xy-xz)(-xz+yz+z²-xy-y²-yz-x²+xy+xz)

=[z(x+y-z)+y(x+y-z)+x(x+y-z)][z(z+y-x)-y(z+y-x)+x(z+y-x)]

=(x+y-z)(x+y+z)(z-y+x)(z+y-x)

i,x³-8-(x-2)(x+4)

=x³-2³-(x-2)(x+4)

=(x-2)(x²+2x+4)-(x-2)(x+4)

=(x-2)[(x²+2x+4)-(x+4)]

=(x-2)[x²+2x+4-x-4]

=(x-2)[x²+x]

=x(x+1)(x+2)