Question

Ajutor la 4 si la 6 va rog

Answer 1

4)
a=(3√2-2√3)(3√2-2√3)-√72-(3-√2)²=
=18-12√6+12-6√2-9+6√2-2=
=18-12√6+11=7-12√6
in carte este o eroare (la a 2-a paranteza este - in loc de +)

ca sa dea numar intreg :
a=(3√2-2√3)(3√2+2√3)-√72-(3-√2)²=
=18-12-6√2-9+6√2-2=
=18-12-9-2=-5  ∈Z   

6)
E(x)=x/(x³+x²) :[(x+2)(2x-1)-x(x+3)+1]/(2x+2)(3x-3)=
=1/x(x+1) × 6(x+1)(x-1)/ (2x²+4x-x-2-x²-3x+1)=
=1/x × 6(x-1)/ (x²-1)=
=1/x × 6/(x+1)=
= 6/ x(x+1)