Given:
The equation is
[tex]\dfrac{???}{x-2}/\dfrac{x^2-1}{x^2-4x+4}=\dfrac{5(x-2)}{x-1}[/tex]
To find:
The missing value.
Solution:
Let the missing value be k.
[tex]\dfrac{k}{x-2}/\dfrac{x^2-1}{x^2-4x+4}=\dfrac{5(x-2)}{x-1}[/tex]
[tex]\dfrac{k}{x-2}\times \dfrac{x^2-2(x)(2)+2^2}{x^2-1^2}=\dfrac{5(x-2)}{x-1}[/tex]
Using the formulae [tex](a-b)^2=a^2-2ab+b^2[/tex] and [tex](a-b)(a+b)=a^2-b^2[/tex].
[tex]\dfrac{k}{x-2}\times \dfrac{(x-2)^2}{(x-1)(x+1)}=\dfrac{5(x-2)}{x-1}[/tex]
[tex]\dfrac{k(x-2)}{(x-1)(x+1)}=\dfrac{5(x-2)}{x-1}[/tex]
Cancel out the common factors from both sides.
[tex]\dfrac{k}{x+1}=5[/tex]
Multiply both sides by (x+1).
[tex]k=5(x+1)[/tex]
Therefore, the missing value is 5(x+1).