"Spoooooooky" Russian rational expressions 2

Algebra Level 3

E = a b + b 2 3 b 3 3 a + a + b b \mathscr{E} = \sqrt{ \dfrac{ \dfrac{ab + b^2}{3} }{\dfrac{b^3}{3a}} + \dfrac{a+b}{b} }

If a = 2016 a = 2016 and b = 2017 b = 2017 , and E \mathscr{E} can be represented in the form x y \dfrac{x}{y} , where x x and y y are coprime positive integers , find x + y x+y .

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


The answer is 6050.

View wiki
Hobart Pao by Hobart Pao , 23, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Chew-Seong Cheong
Jun 30, 2016

E = a b + b 2 3 b 3 3 a + a + b b = a b + b 2 3 × 3 a b 3 + a + b b = a ( a + b ) b 2 + a + b b = a ( a + b ) + b ( a + b ) b 2 = ( a + b ) 2 b 2 = a + b b = 2016 + 2017 2017 = 4033 2017 \begin{aligned} \mathscr E & = \sqrt{\frac {\frac{ab+b^2}3}{\frac {b^3}{3a}} + \frac{a+b}b} \\ & = \sqrt{\frac{ab+b^2}3 \times \frac {3a}{b^3} + \frac{a+b}b} \\ & = \sqrt{\frac{a(a+b)}{b^2} + \frac{a+b}b} \\ & = \sqrt{\frac{a(a+b)+b(a+b)}{b^2}} \\ & = \sqrt{\frac{(a+b)^2}{b^2}} \\ & = \frac {a+b}b = \frac {2016+2017}{2017} = \frac {4033}{2017} \end{aligned}

⟹ x + y = 4033 + 2017 = 6050 \implies x+y = 4033+2017 = \boxed{6050}

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice , Same Solution +1

Novril Razenda - 4 years, 11 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...