2012 x 2012 isn't huge, right?

Algebra Level 3

1 + 201 1 2 + 2011 2 201 2 2 + 2011 2012 \sqrt { 1+2011^{ 2 }+\frac { { 2011 }^{ 2 } }{ 2012^{ 2 } } } +\frac { 2011 }{ 2012 }

Evaluate the expression above and write your answer to 3 decimal places.


The answer is 2012.000.

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

x = 1 + 201 1 2 + 201 1 2 201 2 2 + 2011 2012 Let a = 2012 = 1 + ( a 1 ) 2 + ( a 1 ) 2 a 2 + a 1 a = 1 + a 2 2 a + 1 + 1 2 a + 1 a 2 + 1 1 a = a 2 + 2 + 1 a 2 2 ( a + 1 a ) + 1 + 1 1 a = ( a + 1 a ) 2 2 ( a + 1 a ) + 1 + 1 1 a = ( a + 1 a 1 ) 2 + 1 1 a = a + 1 a 1 + 1 1 a = a = 2012 \begin{aligned} x & = \sqrt{1+2011^2+\frac {2011^2}{2012^2}} + \frac {2011}{2012} & \small \color{#3D99F6} \text{Let }a = 2012 \\ & = \sqrt{1+(a-1)^2+\frac {(a-1)^2}{a^2}} + \frac {a-1}a \\ & = \sqrt{1+a^2-2a+1+1-\frac 2a + \frac 1{a^2}} + 1 - \frac 1a \\ & = \sqrt{a^2 + 2 + \frac 1{a^2} -2\left(a+\frac 1a\right) + 1} + 1 - \frac 1a \\ & = \sqrt{\left(a+\frac 1a\right)^2 -2\left(a+\frac 1a\right) + 1} + 1 - \frac 1a \\ & = \sqrt{\left(a+\frac 1a - 1\right)^2} + 1 - \frac 1a \\ & = a+\frac 1a - 1 + 1 - \frac 1a \\ & = a = \boxed{2012} \end{aligned}

Nice solution!

Steven Jim - 4 years, 1 month ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...