Tuesday, July 6, 2021

Vietnam Math Out Standing Student 2012-13

 

Vietnam Math Out Standing Student 2012-13

Problem 01: Solve the equation of: x4n+x2n+2012=2012    for all nN.
Problem 02: It is given (Un) which is determined by: 
                    U1=3
                    Un+1=13(2Un+3Un2)    for all nN.
                    Let's finding the limit of: limn(un).
Problem 03: It is given three non-negative real numbers x,y,z , Prove that:
                    1x+1y+1z>369+x2y2+y2^2+z2x2.
Problem 04: Find roots positive numbers of equation:
                    x+23=y+z

Solution

Problem 01: We had the equation: x4n+x2n+2012=2012    for all nN.
                    Let's t=x2n and replace into our equation above:
                    t2+t+2012=2012
                      t2+t+14=t+2012-t+2012+14
                      (t+12)2=(t+2012-12)2
                     t+1=t+2012
                       t2+t-2011=0         (1)
Solve the equation (1) we got: t=-1+80452
Then, we replace the value of t in tot=x2n we will get the roots of our problem.



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