Question

Menționați condițiile în care fracțiile date sunt definite și simplificați-le:
a) x²+5x+6/(x+2)²=
b)(x-3)²/x²-7x+12=
c)(x+2)²/x²+3x+2=
d)x²-3x+2/x²-4x+3=
e)x²+5x+4/x²+6x+8=
f)x²+3x-4/x²+5x-6=
g)x²+x-2/x²-3x-10=
h)x²-x-6/x²+2x-15=
i)x²+2x-24/x²+4x-12=
j)x²-4x-21/x²-x-12=
k)x²+7x-30/x²+8x-20=
l)x²-4x-5/x²-11x+30=

Answer 1

a) x²+5x+6/(x+2)²=
=x²+2x+3x+6/(x+2)²=
=x(x+2)+3(x+2)/(x+2)²=
=(x+2)(x+3)/(x+2)²=
=(x+3)/(x+2)
x≠-2

b)(x-3)²/x²-7x+12=
=(x-3)²/x²-4x-3x+12=
=(x-3)²/x(x-4)-3(x-4)=
=(x-3)²/(x-4)(x-3)=                              x≠4    x≠3
=(x-3)/(x-4)

 c)(x+2)²/x²+3x+2=
=(x+2)²/x²+x+2x+2=
=(x+2)²/x(x+1)+2(x+1)=
=(x+2)²/(x+1)(x+2)=                    x≠-1    x≠-2
=(x+2)/(x+1)


d)x²-3x+2/x²-4x+3=
=x²-x-2x+2/x²-x-3x+3=
=x(x-1)-2(x-1)/x(x-1)-3(x-1)=
=(x-1)(x-2)/(x-1)(x-3)=                x≠1  x≠3
=(x-2)/(x-3)

e)x²+5x+4/x²+6x+8=
=(x+1)(x+4)/(x+2)(x+4)=                      x≠-2  x≠-4
=(x+1)/(x+4)=

f)x²+3x-4/x²+5x-6=
=(x+4)(x-1)/(x+6)(x-1)=                          x≠-6  x≠1
=(x+4)/(x+6)

 g)x²+x-2/x²-3x-10=
=(x+2)(x-1)/(x-5)(x+2)=                        x≠5    x≠-2
=(x-1)/(x-5)=
 
h)x²-x-6/x²+2x-15=
=(x-3)(x+2)/(x+5)(x-3)=                  x≠-5  x≠3
=(x+2)/(x+5)

i)x²+2x-24/x²+4x-12=
=(x+6)(x-4)/(x+6)(x-2)=          x≠-6  x≠2
=(x-4)/(x-2)

j)x²-4x-21/x²-x-12=
=(x-7)(x+3)/(x-4)(x+3)=              x≠4    x≠-3
=(x-7)/(x-4)=

k)x²+7x-30/x²+8x-20=
=(x+10)(x-3)/(x+10)(x-2)=        x≠-10  x≠2
=(x-3)/(x-2)

l)x²-4x-5/x²-11x+30=
=(x-5)(x+1)/(x-5)(x-6)=              x≠5    x≠6
=(x+1)/(x-6)