Răspuns:
Explicație pas cu pas:
Vom descompune in factori numitorii fractiilor din paranteza rotunda
x²-x=x(x-1); x²-1²=(x-1)(x+1); x²+x=x(x+1).
Atunci numitorul comun este x(x-1)(x+1)
Facem operatiile din paranteze
[tex]\frac{x+2}{x(x-1)}+\frac{x-3}{(x-1)(x+1)}-\frac{x-1}{x(x+1)} = \frac{(x+2)(x+1)+x(x-3)-(x-1)(x-1)}{x(x-1)(x+1)}=\frac{x^{2}+3x+2+x^{2}-3x-x^{2}+2x-1}{x(x-1)(x+1)}=\frac{x^{2}+2x+1}{x(x-1)(x+1)} =\frac{(x+1)^{2}}{x(x-1)(x+1)}=\frac{x+1}{x(x-1)}.\\[/tex]
Descompunem in factori numaratorul si numitorul ultimei fractii.
x²+3x+2=x²+2x+1+x+1=(x+1)²+(x+1)=(x+1)(x+1+1)=(x+1)(x+2).
x³-2x²+x=x(x²-2x+1)=x(x-1)². Acum facem ultima operatie
[tex]\frac{x+1}{x(x-1)}:\frac{(x+1)(x+2)}{x(x+1)^{2}} =\frac{x+1}{x(x-1)}*\frac{x(x-1)^{2}}{(x+2)(x+1)}=\frac{x-1}{x+2} .\\Deci~E(x)=\frac{x-1}{x+2} .\\b)~E(x)=\frac{3(x+1)}{(x+2)} ,~\frac{x-1}{x+2}=\frac{3(x+1)}{(x+2)} ,~deci~x-1=3(x+1)\\[/tex]
x-1=3x+3, ⇒x-3x=3+1, ⇒-2x=4, ⇒x=4:(-2)=-2.
