TA có: \(\left|x-3y\right|^{2007}\ge0\forall x,y\)

\(\left|y+4\right|^{2008}\ge0\forall y\)

Do đó: \(\left|x-3y\right|^{2007}+\left|y+4\right|^{2008}\ge0\forall x,y\)

mà \(\left|x-3y\right|^{2007}+\left|y+4\right|^{2008}<=0\)

nên x-3y=0 và y+4=0

=>y=-4 và x=3y=-12

Kết quả bằng 2009

Xét tử

2008+2007/2+2006/3+2005/4+ ... +2/2007+1/2008

=(1+1+1+...+1)+2007/2+2006/3+2005/4+ ... +2/2007+1/2008

= 1+ (2007/2)+1+(2006/3)+1+(2005/4)+1+ ... + (2/2007)+1+(1/2008)+1

=2009/2009+2009/2+2009/3+2009/4+ ... + 2009/2007 + 2009/2008

=2009.(1/2+1/3+1/4+ ... + 1/2007+1/2008+1/2009)

\(B=2008+\frac{2007}{2}+\frac{2006}{3}+\frac{2005}{4}+...+\frac{2}{2007}+\frac{1}{2008}\)

\(=1+1+\frac{2007}{2}+1+\frac{2006}{3}+...+1+\frac{1}{2008}\)

\(=\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2008}\)

\(=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\)

Suy ra \(A=2009\).

A=2008+2007/2+2006/3+2005/4+...+2/2007+1/2008

1/2+1/3+1/4+1/5+...+1/2007+1/2008

=(1+2007/2)+(1+2006/3)+(1+2005/4)+...+(1+2/2007)+(1+1/2008)

1/2+1/3+1/4+...+1/2008

=2009(1/2+1/3+1/4+...+1/2008)

1/2+1/3+1/4+..+1/2008

=2009

CÁi thứ hai :

   vì 2007 (2x -y)^2008 >= 0 để bt <0 => 2x - y = 0 => 2x=y 

 => y- 4 = 0 => y = 4 

2x = 4 => x = 2 

VẬy x = 2 ;4 

Thứ ba :

   Vì 2( x- 5)^4 >= 0 

Để 2( x- 5)^4 + 5(2y- 7)^5 = 0 

= > x- 5 = 0 => x = 5

2y -7 = 0 => y = 7/2