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Sauron [17]
3 years ago
14

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

Mathematics
1 answer:
o-na [289]3 years ago
5 0

Given:

The equation is

\dfrac{???}{x-2}/\dfrac{x^2-1}{x^2-4x+4}=\dfrac{5(x-2)}{x-1}

To find:

The missing value.

Solution:

Let the missing value be k.

\dfrac{k}{x-2}/\dfrac{x^2-1}{x^2-4x+4}=\dfrac{5(x-2)}{x-1}

\dfrac{k}{x-2}\times \dfrac{x^2-2(x)(2)+2^2}{x^2-1^2}=\dfrac{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.

\dfrac{k}{x-2}\times \dfrac{(x-2)^2}{(x-1)(x+1)}=\dfrac{5(x-2)}{x-1}

\dfrac{k(x-2)}{(x-1)(x+1)}=\dfrac{5(x-2)}{x-1}

Cancel out the common factors from both sides.

\dfrac{k}{x+1}=5

Multiply both sides by (x+1).

k=5(x+1)

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

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