👤
Smallworldsanalolee
a fost răspuns

Se da E(x) = am pus poza ....
a) Aratati ca E(x) = 1 + [tex] \frac{1}{x-2} [/tex]
b) Det x ∈ Z pt care E(x) ∈ Z

Se Da Ex Am Pus Poza A Aratati Ca Ex 1 Tex Frac1x2 Tex B Det X Z Pt Care Ex Z class=

Răspuns :

Matepentrutoti
[tex]a)E(x)=\frac{x^2-2x+1}{(x-2)^2} * \frac{x-2}{x-1} =\\ =\frac{(x-1)^2}{(x-2)^2} * \frac{x-2}{x-1}= \frac{x-1}{x-2} =\\ =\frac{x-2+1}{x-2} =1+ \frac{1}{x-2} [/tex]
b) E(x) ∈ Z=>1/(x-2)∈ Z=>
x-2=1=>x=3
x-2=-1=>x=1
Cioroiuroxana
x²-4x+4= (x-2)²

x/(x-2 )+ 1/(x²-4x+4) = x/(x-2 )+ 1/(x-2)²= (x*(x-2)+1)/ (x-2)²= x²-2x+1/(x-2)²=
= (x-1)²/ (x-2)²
=>   x/(x-2 )+ 1/(x²-4x+4)  :  (x-1)/(x-2) = (x-1)²/ (x-2)² * (x-2)/(x-1)= (x-1)/(x-2)= (x-2+1)/(x-2) = 1+ 1/x-2