"Spoooooooky" Russian rational expressions 6

Algebra Level 2

E = ( a b 2 − a b + b a 2 − a b ) ( a b b − a ) \mathscr{E} = \left( \dfrac{a}{b^2 - ab} + \dfrac{b}{a^2 -ab} \right) \left( \dfrac{ab}{b-a} \right)

Let a = 2017 a = 2017 and b = 2016 b = 2016 . Find the exact value of E \mathscr{E} .

Credit: My former Trig teacher's worksheet of Russian rational expressions


The answer is 4033.

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Hobart Pao by Hobart Pao , 23, USA

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1 solution

⇒ ( a b 2 − a b + b a 2 − a b ) ( a b b − a ) \Rightarrow \left( \dfrac{a}{b^2 - ab} + \dfrac{b}{a^2 -ab} \right) \left( \dfrac{ab}{b-a} \right)

= ( a b ( b − a ) − b a ( b − a ) ) × a b b − a =\left(\dfrac{a}{b(b-a)}-\dfrac{b}{a(b-a)}\right)×\dfrac{ab}{b-a}

= 1 b − a ( a b − b a ) × a b b − a =\dfrac{1}{b-a}\left(\dfrac{a}{b}-\dfrac{b}{a}\right)×\dfrac{ab}{b-a}

= 1 b − a ( a 2 − b 2 a b ) × a b b − a =\dfrac{1}{b-a}\left(\dfrac{a^2-b^2}{ab}\right)×\dfrac{ab}{b-a}

= − 1 a − b × ( a + b ) ( a − b ) × 1 b − a =\dfrac{-1}{a-b}×(a+b)(a-b)×\dfrac{1}{b-a}

= − ( a + b ) b − a =\dfrac{-(a+b)}{b-a}

Pluging a = 2017 a=2017 and b = 2016 b=2016 .

− ( 4033 ) − 1 = 4033 \dfrac{-(4033)}{-1}=\boxed{4033}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Alternatively, you get a + b a − b \dfrac{a+b}{a-b} , eliminating the need for negatives in that last step but good job!

Hobart Pao - 4 years, 11 months ago

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Thanks! :)

A Former Brilliant Member - 4 years, 11 months ago

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