Question

Am atasat poza vreu exercitiul 5 multumesc

Answer 1

;Răspuns:

Explicație pas cu pas:

E(x)=(2x+1)/(x-1):(1-4x²)/(1-x²)=(2x+1)/(x-1)×(x-1)(x+1)/(2x-1)(2x+1)=

=(x+1)/(2x-1)   x≠±1   x≠±1/2

Answer 2

Răspuns:

Explicație pas cu pas:

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

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

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

$\displaystyle E(x)=\frac{2x+1}{x-1} :\frac{1-4x^{2} }{1-x^{2} } \\ \\ \displaystyle E(x)=\frac{2x+1}{x-1} \times\frac{1-x^{2} }{1-4x^{2} } \\ \\ \displaystyle E(x)=\frac{\not (2x+1)}{x-1} \times \frac{(1-x)(1+x)}{(1-2x)\not(1+2x)}$

$\displaystyle E(x)=\frac{(1-x)(1+x)}{(x-1)(1-2x)} \\ \\ \displaystyle E(x)=\frac{\not(1-x)(1+x)}{-\not(1-x)(1-2x)} \\ \\ \displaystyle E(x)=\frac{1+x}{-(1-2x)} \\ \\ \displaystyle E(x)=\frac{x+1}{2x-1}$

$\displaystyle x^{2} -y^{2} =(x-y)(x+y)$

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