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Efectuati: [tex]a)) \frac{2-x}{3x+12} :\frac{x^{2}-4}{4x+x^{2}}
b)) \frac{3aab}{ax+3x}: \frac{6ba^{2}}{9+6a+a^{2}} [/tex]

Răspuns :

Renatemambouko
a) (2-x)/(3(x+4) : (x-2)(x+2)/ x(4+x)=
=(2-x)/3(x+4) :  x(4+x)/(x-2)(x+2)=
=-(x-2)/3 :  x/(x-2)(x+2)=
=-1/3 :  x/(x+2)=
=-x/3(x+2)

b)
3aab/x(a+3): 6ba²/(3+a)²=
=3a²b/x(a+3) × (3+a)²/6ba²=
=1/x × (3+a)/2=
=(a+3)/2x



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