+0  
 
0
885
5
avatar+18 

(2001^2 - 1992^2) / (2003^2 - 1997^2) = ?

 Jul 16, 2016

Best Answer 

 #1
avatar+33666 
+20

Use the fact that a^2 - b^2 = (a+b)(a-b) so we have:  (2001+1992)*9/[(2003+1997)*6]

 

or 3993*9/(4000*6) ≈ 1.497

 

.

 Jul 16, 2016
 #1
avatar+33666 
+20
Best Answer

Use the fact that a^2 - b^2 = (a+b)(a-b) so we have:  (2001+1992)*9/[(2003+1997)*6]

 

or 3993*9/(4000*6) ≈ 1.497

 

.

Alan Jul 16, 2016
 #5
avatar+67 
0

I got the same answer

MichaelShaun  Jul 17, 2016
 #2
avatar+118724 
+10

Good thinking Alan :)

 

Here is another way - I used the calculator.  LOL

 

(2001^2 - 1992^2) / (2003^2 - 1997^2) = 1.497375

 Jul 16, 2016
 #3
avatar+18 
0

actually sorry guys my bad... it should have been   " (2001^2 - 1999^2) / (2003^2 - 1997^2) = ? "

i coulnt notice the "a^2 - b^2" fact, thanks for the answer Alan and who tried to help me :)

with the a^2 - b^2 fact i easily found the answer of my question it is 1/3 :) 

and im really sory for my mistake, thank you all for helping out :)

 Jul 16, 2016
 #4
avatar+9676 
0

\(\boxed{a^2-b^2=(a+b)(a-b)}\\\frac{2001^2-1999^2}{2003^2-1997^2}= \frac{(2001+1999)(2001-1999)}{(2003+1997)(2003-1997)} \\\;\;\;\;\;\qquad\quad=\dfrac{4000\times 2}{4000 \times 6}=\dfrac{2}6=\dfrac{1}3\)

MaxWong  Jul 17, 2016

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