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What is the missing expression that would make the left-hand-side of the equation equal to the right-hand-side?


(???)/(x-2)/(x^(2)-1 )/(x^(2) -4x+4) =(5(x-2))/(x-1)

User H Ketabi
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1 Answer

2 votes

Given:

The equation is


(???)/(x-2)/(x^2-1)/(x^2-4x+4)=(5(x-2))/(x-1)

To find:

The missing value.

Solution:

Let the missing value be k.


(k)/(x-2)/(x^2-1)/(x^2-4x+4)=(5(x-2))/(x-1)


(k)/(x-2)* (x^2-2(x)(2)+2^2)/(x^2-1^2)=(5(x-2))/(x-1)

Using the formulae
(a-b)^2=a^2-2ab+b^2 and
(a-b)(a+b)=a^2-b^2.


(k)/(x-2)* ((x-2)^2)/((x-1)(x+1))=(5(x-2))/(x-1)


(k(x-2))/((x-1)(x+1))=(5(x-2))/(x-1)

Cancel out the common factors from both sides.


(k)/(x+1)=5

Multiply both sides by (x+1).


k=5(x+1)

Therefore, the missing value is 5(x+1).

User Benny Chan
by
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