Question

ma poate ajuta cineva la 30???​

Answer 1

Răspuns:

Explicație pas cu pas:

30

a) x²-8x+15=x²-5x-3x+15=x(x-5)-3(x-5)=(x-5)·(x-3)=(x-3)·(x-5)

b) E(x)=[(x+2)/(x-5)-14/(x²-8x+15)]:1/(x²-9)=

           [(x+2)/(x-5)-14/(x-3)·(x-5)]:1/(x+3)·(x-3)=

           [(x+2)·(x-3)-14]/[(x-3)·(x-5)]·[(x+3)·(x-3)]=

           [(x²+2x-3x-6-14)/(x-3)·(x-5)]·[(x+3)·(x-3)]=

           [(x²-x-20)/(x-3)·(x-5)]·[(x+3)·(x-3)]=

           [(x²-16-x-4)·(x+3)·(x-3)]/[(x-3)·(x-5)]=

           [(x+4)·(x-4)-(x+4)]·[(x+3)·(x-3)]/[(x-3)·(x-5)]=

           [(x+4)·(x-4-1)]·[(x+3)·(x-3)]/[(x-3)·(x-5)]=

           [(x+4)·(x-5)·(x+3)·(x-3)]/[(x-3)·(x-5)]=

           [(x+4)·(x+3)·(x-3)]/[(x-3)]=

            (x+4)·(x+3)=(x+3)(x+4)