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Aratati ca expresia E(x)=[tex] \frac{(x+2)^{2} +(2x-1)^2}{(x-3)^2+(3x+1)^2} [/tex] are valoare constanta pentru orice numar real x.

Răspuns :

[(x+2)²+(2x-1)²]/[(x-3)²+(3x+1)²]=
=(x²+4x+4+4x²-4x+1)/(x²-6x+9+9x²+6x+1)=
=(x²+4+4x²+1)/(x²+9+9x²+1)=
=(5x²+5)/(10x²+10)=
=5(x²+1)/10(x²+1)=5/10=1/2

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