Respuesta :
Answer:
The simplified form is [tex]\dfrac{2(x-1)}{3(x+1)}[/tex].
[tex]x =1[/tex] is the excluded value for the given expression.
Step-by-step explanation:
Given:
The expression given is:
[tex]\dfrac{2a^2-4a+2}{3a^2-3}[/tex]
Let us simplify the numerator and denominator separately.
The numerator is given as [tex]2a^2-4a+2[/tex]
2 is a common factor in all the three terms. So, we factor it out. This gives,
[tex]=2(a^2-2a+1)[/tex]
Now, [tex]a^2-2a+1=(a-1)(a-1)[/tex]
Therefore, the numerator becomes [tex]2(a-1)(a-1)[/tex]
The denominator is given as: [tex]3a^2-3[/tex]
Factoring out 3, we get
[tex]3(a^2-1)[/tex]
Now, [tex]a^2-1[/tex] is of the form [tex]a^2-b^2=(a-b)(a+b)[/tex]
So, [tex]a^2-1=(a-1)(a+1)[/tex]
Therefore, the denominator becomes [tex]3(a-1)(a+1)[/tex]
Now, the given expression is simplified to:
[tex]\frac{2a^2-4a+2}{3a^2-3}=\frac{2(x-1)(x-1)}{3(x-1)(x+1)}[/tex]
There is [tex](x-1)[/tex] in the numerator and denominator. We can cancel them only if [tex]x\ne1[/tex] as for [tex]x=1[/tex], the given expression is undefined.
Now, cancelling the like terms considering [tex]x\ne1[/tex], we get:
[tex]\dfrac{2a^2-4a+2}{3a^2-3}=\dfrac{2(x-1)}{3(x+1)}[/tex]
Therefore, the simplified form is [tex]\dfrac{2(x-1)}{3(x+1)}[/tex]
The simplification is true only if [tex]x\ne1[/tex]. So, [tex]x =1[/tex] is the excluded value for the given expression.