Question

Rezolvati exercitiul din poza 100 puncte urgent

Answer 1

Răspuns:

Explicație pas cu pas:

$E(x)=(\frac{x+1}{x-3}-\frac{x^{2}+3x+2}{x^{2}+4x+3}-\frac{1}{9-x^{2}}) :\frac{x+2}{x^{2}-9}   =(\frac{x+1}{x-3}-\frac{x^{2}+2x+1+x+1}{x^{2}+2x+1+2x+2} +\frac{1}{x^{2}-9}) :\frac{x+2}{x^{2}-9}= (\frac{x+1}{x-3}-\frac{(x+1)^{2}+(x+1)}{(x+1)^{2}+2(x+1)}+\frac{1}{(x-3)(x+3)}) :\frac{x+2}{x^{2}-9}=(\frac{x+1}{x-3}-\frac{(x+1)(x+1+1)}{(x+1)(x+1+2)}+ \frac{1}{(x-3)(x+3)}) :(\frac{x+1}{x-3}- \frac{x+2}{x+3}+ \frac{1}{(x-3)(x+3)}) :\frac{x+2}{x^{2}-9}=$

$= \frac{(x+1)(x+3)-(x+2)(x-3)+1}{x^{2}-9} : \frac{x+2}{x^{2}-9}=\frac{x^{2}+4x+3-x^{2}+x+6+1}{x^{2}-9}*\frac{x^{2}-9}{x+2}=\frac{5x+10}{x^{2}-9} *\frac{x^{2}-9}{x+2}= \frac{5(x+2)}{x+2} =5.$

Deci E(x)=5 pentru orice x din Domeniul de definitie.

E(m)=2m+1=5, ⇒2m=5-1, ⇒2m=4, ⇒m=2.