Question

Ma puteți ajuta va rog la exercițile de la problema 4?​

Answer 1

Explicație pas cu pas:

(x-4)(2x+1)²-1(x-4)=(x-4)*[(2x+1)²-1]=(x-4)(2x+1-1)(2x+1+1)=

2x(x-4)(2x+2)=4x(x-4)(x+1)

(4x-3)²-9(x+2)²=[(4x-3)-3(x+2)]*[(4x-3)+3(x+2)]=

(4x-3-3x-6)(4x-3+3x+6)=(x-9)(7x+3)

Answer 2

Răspuns:

a)

(x-4)(2x+1)^2-x+4=

= (x-4)(2x+1)^2-1(x-4)

scoatem (x-4) factor comun

= (x-4)[(2x+1)^2-1]

avem a^2-b^2= (a+b)(a-b)

= (x-4)(2x+1+1)(2x+1-1)=

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

= 2*2x(x-4)(x+1) (am scos 2 factor comun)

=4x(x-4)(x+1)

b)

(4x-3)^2-9(x+2)^2=

=(4x-3)^2-3^2*(x+2)^2=

=(4x-3)^2-[(3(x+2)^2]

avem a^2-b^2 din nou

= [(4x-3)+(3x+6)][(4x-3)-(3x+6)]=

= (7x+3)(x-9)