Question

Îmi poate explica cineva cum se rezolva ex. astea...

Answer 1

Inmultesti fiecare nr. din prima paranteza cu a doua paranteza:

A)    (x+1)(x+2)=x(x+2)+1(x+2)=x²+2x+x+2=x²+3x+2

B)    (x-3)(x+7)=x(x+7)-3(x+7)=x²+7x-3x-21=x²+4x-21

C)    (2x+5)(x-3)=2x(x-3)+5(x-3)=2x²-6x+5x-15=2x²-x-15

D)    (1-2x)(3x+4)=1·(3x+4)-2x(3x+4)=3x+4-6x²-8x= -6x²-5x+4

E)    (a+3)(5a-2)=a(5a-2)+3(5a-2)=5a²-2a+15a-6=5a²+13a-6

F)     (x+y)(2x-y)=x(2x-y)+y(2x-y)=2x²-xy+2xy-y²=2x²+xy-y²

G)    (2x-7y)(3y+5x)=2x(3y+5x)-7y(3y+5x)=6xy+10x²-21y²-35xy=
        =10x²-29xy-21y²

H)    (3m+2n)(3m-2n)=(3m)²-(2n)²=9m²-4n²  ,
         deoarece e formula (x-y)(x+y)=x²-y²

I)      (m+n)(m²+n²)=m(m²+n²)+n(m²+n²)=m³+mn²+nm²+n³

                           Iaca asa se face "Deschiderea Parantezelor" !