Question

Pentru orice x e R notam E(x)=(x+3)la a 2-a +2(x-4)(x+3)+(x-4) la a 2-a.
a)Sa se arate ca E(x)>0, pentru orice x e R
b)Sa se rezolve in multimea numerelor reale ecuatia E(x)=1

Answer 1

a)E=(x+3)(x+3)+2(x-4)(x+3)+(x-4)(x-4)
E=x*x+6x+9+2(x*x+3x-4x-12)+x*x-8x+16
E=2x*x-2x+25+2x*x+6x-8x-24
E=4x*x-4x+1 <=> (2x-1)(2x-1)

b)E=1=>(2*1-1)(2*1-1)=1