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Bunuel
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The above solution made no sense to me.

Here's one:

\(\frac{a}{b}\) + \(\frac{b}{a}\) - ab

\(\frac{a^2 + b^2 - a^2b^2}{ab}\)

\(\frac{(a-b)^2 + 2ab - (ab)(ab)}{ab}\)

\(\frac{(ab)^2 + 2ab - (ab)(ab)}{ab}\)

ab + 2 - ab

= 2
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The above solution made no sense to me.

Here's one:

\(\frac{a}{b}\) + \(\frac{b}{a}\) - ab

\(\frac{a^2 + b^2 - a^2b^2}{ab}\)

\(\frac{(a-b)^2 + 2ab - (ab)(ab)}{ab}\)

\(\frac{(ab)^2 <}{span>+ 2ab - (ab)(ab)/ab}\)

ab + 2 - ab

= 2


Can you please explain how did you derive the underlined portion?
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The above solution made no sense to me.

Here's one:

\(\frac{a}{b}\) + \(\frac{b}{a}\) - ab

\(\frac{a^2 + b^2 - a^2b^2}{ab}\)

\(\frac{(a-b)^2 + 2ab - (ab)(ab)}{ab}\)

\(\frac{(ab)^2 <}{span>+ 2ab - (ab)(ab)/ab}\)

ab + 2 - ab

= 2


Can you please explain how did you derive the underlined portion?

I canceled out the (ab) in each term.

\(\frac{(ab)^2 + 2ab - (ab)(ab)}{ab}\)

Since we have (ab) in the denominator, we can cancel that out, and also cancel out the (ab) in each term in the numerator. But we can't cancel it out twice if a term has it twice, we can only cancel it out once.

\(\frac{(ab)(ab) + 2(ab) - (ab)(ab)}{ab}\)

So that becomes \((ab) + 2 - (ab)\)

Does that make sense?
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I think about two number that satisfy ab=a-b , so a=1 and b =1/2
then substitute into question

answer is 2, E
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ab = a -b

Squaring both the sides we get,

\((ab)^2 = (a - b)^2\)

=> \(a^2b^2 = a^2 + b^2 - 2ab\)

=> \(\frac{a}{b} + \frac{b}{a} - ab: \frac{(a^2 + b^2 - a^2b^2)}{ab}\)

=> \(\frac{a}{b} + \frac{b}{a} - ab: \frac{(a^2 + b^2 - [a^2 + b^2 - 2ab])}{ab}\)

=> \(\frac{a}{b} + \frac{b}{a} - ab: \frac{2ab}{ab}\)

=> \(\frac{a}{b} + \frac{b}{a} - ab: 2\)

Answer E
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