Răspuns:
Explicație pas cu pas:
[tex]\displaystyle E(x) =\frac{2x+1}{x-1} : (1-\frac{3x^{2} }{1-x^{2} })[/tex]
[tex]\displaystyle E(x)=\frac{2x+1}{x-1} :(\frac{1-x^{2} }{1-x^{2} } -\frac{3x^{2} }{1-x^{2} } )[/tex]
[tex]\displaystyle E(x)=\frac{2x+1}{x-1} :\frac{1-x^{2} -3x^{2} }{1-x^{2} }[/tex]
[tex]\displaystyle E(x)=\frac{2x+1}{x-1} :\frac{1-4x^{2} }{1-x^{2} } \\ \\ \displaystyle E(x)=\frac{2x+1}{x-1} \times\frac{1-x^{2} }{1-4x^{2} } \\ \\ \displaystyle E(x)=\frac{\not (2x+1)}{x-1} \times \frac{(1-x)(1+x)}{(1-2x)\not(1+2x)}[/tex]
[tex]\displaystyle E(x)=\frac{(1-x)(1+x)}{(x-1)(1-2x)} \\ \\ \displaystyle E(x)=\frac{\not(1-x)(1+x)}{-\not(1-x)(1-2x)} \\ \\ \displaystyle E(x)=\frac{1+x}{-(1-2x)} \\ \\ \displaystyle E(x)=\frac{x+1}{2x-1}[/tex]
[tex]\displaystyle x^{2} -y^{2} =(x-y)(x+y)[/tex]
punem si conditiile de existentialitate
x-1≠0⇒x≠1
x+1≠0⇒ x≠-1
2x-1≠0⇒x≠1/2
2x+1≠0⇒x≠-1/2
