thoa man de bai 21 < x < 83 nen x = 22 ;23 ;24 ;25 ;26 ;27 ;28 ,29 , 30 , ..................... 83 

=> A ={ 22 ; 23 ; 24 ; 25 ; .............. 83 }

211 con vịt 

211 con

1 face

2 big mirror

3 Alice

4 black lines

5 moustache and a beard

6 a green face

7 scary

Với mọi x;y;z ta luôn có:

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)

\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\)

\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)\ge3\left(xy+yz+zx\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Rightarrow xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}=\frac{3^2}{3}=3\)

\(B_{max}=3\) khi \(x=y=z=1\)

\(f\left(x+1\right)=\left(x+1\right)^2+p\left(x+1\right)+q\)

\(=x^2+2x+1+px+p+q\)

\(=\left(x^2+px+q\right)+2x+q+1=f\left(x\right)+2x+p+1\)

\(\Rightarrow f\left(x\right).f\left(x+1\right)=f\left(x\right)\left[f\left(x\right)+2x+p+1\right]\)

\(=f^2\left(x\right)+2x.f\left(x\right)+x^2-x^2+p.f\left(x\right)+f\left(x\right)\)

\(=\left[f\left(x\right)+x\right]^2-x^2+p.f\left(x\right)+x^2+px+q\)

\(=\left[f\left(x\right)+x\right]^2+p\left[f\left(x\right)+x\right]+q\)

\(=f\left[f\left(x\right)+x\right]\)

Thay \(x=2008\Rightarrow f\left(2008\right).f\left(2009\right)=f\left[f\left(2008\right)+2008\right]\)

Vậy tồn tại số nguyên k sao cho \(f=f\left(2008\right)+2008\) để \(f\left(k\right)=f\left(2008\right).f\left(2009\right)\)

b b b

B

B 

C

Chọn B