Question

Ma poate ajuta cineva, va rog?​

Answer 1

Răspuns:

2.a)(x²+2x)( x²+2x+5)+6=(x²+2x+2)(x²+2x+3)

(x²+2x)²+5x²+10x+6=

(x²+2x)²+5(x²+2x)+6=

(x²+2x)²+4(x²+2x)+4+x²+2x+2=

(x²+2x+2)²+x²+2x+2=

(x²+2x+2)(x²+2x+2+1)=

(x²+2x+2)(x²+2x+3)

b)E(x)=[(x²+2x+2)²-1]/(x²+2x+2)(x²+2x+3)=

[(x²+2x+2-1)(x²+2x+3)]/

(x²+2x+2)(x²+2x+3)=(x²+2x+1)/(x²+2x+2)=>

E(x)=1+1/(x+1)²val minimă

Kim când x=>∞(x+1)²max

1/(x+1)²=0

E(x)=1

3.a)f(x)=x-1

f(a)+f(-a)+2=a-1+-a-1+2=0+0=0

b)OP=5

x²p+y²p=OP²=25

x aparține funcției x+1

x²+(x+1)²=25

x²+x²+2x+1=25

2x²+2x-24=0/:2

x²+x-12=0

x1=-4. x2=3

y1=-4-1=-5

y2=3-1=2

P1(-4;-5). P2(3;2)