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Beatriceionela
a fost răspuns

E(x)=[tex]( \frac{x+3}{ x^{2} +6x+9} - \frac{x}{x+3} ) * \frac{x+3}{-3x-3x+6} [/tex]

Răspuns :

Renatemambouko
E(x)=[(x+3)/(x²+6x+9)-x/(x+3)] × (x+3)/(-3x²-3x+6)=
=[(x+3)/(x+3)²-x/(x+3)] × (x+3)/(-3)(x²+x-2)=
=[1/(x+3)-x/(x+3)] × (x+3)/(-3)(x-1)(x+2)=
=(1-x)/(x+3)] × (x+3)/3(1-x)(x+2)=
=1/3(x+2)




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