Ta có :\(\left(a-b\right)^2\ge0\forall a;b\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow2a^2+2b^2\ge a^2+b^2+2ab\Leftrightarrow2a^2+2b^2\ge\left(a+b\right)^2\)

Suy ra \(\frac{2011}{2a^2+2b^2+2008}\le\frac{2011}{\left(a+b\right)^2+2008}=\frac{2011}{4+2008}=\frac{2011}{2012}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=1\)

Thiếu đề k bn ???

Đặt \(Q=\dfrac{2011}{2a^2+2b^2+2008}\)

Ta có:

\(\dfrac{a+b}{2}=1=>a+b=2=>a=2-b\)

Thay a=2-b vào Q ta được:

\(Q=\dfrac{2011}{2a^2+2\left(2-a\right)^2+2008}\)

=\(\dfrac{2011}{2a^2+2\left(4-4a+a^2\right)+2008}\)

=\(\dfrac{2011}{2a^2+8-8a+2a^2+2008}\)

=\(\dfrac{2011}{4a^2-8a+2016}\)

=\(\dfrac{2011}{4a^2-8a+4+2012}\)

=\(\dfrac{2011}{4\left(a^2-2a+1\right)+2012}\)

=\(\dfrac{2011}{4\left(a-1\right)^2+2012}\)

Vì \(2a^2+2b^2+2008>0với\forall a,b\)

nên để Q đạt GTLN thì \(2a^2+2b^2+2008\)đạt GTNN hay \(4\left(a-1\right)^2+2012\)đạt GTNN

Mặt khác \(4\left(a-1\right)^2\)\(\ge\)0 với \(\forall\)a

Do đó\(4\left(a-1\right)^2+2012\) \(\ge\)0 với \(\forall\)a

Dấu "=" xảy ra <=> a-1=0<=>a=1

Mà a+b=2=>b=1

Vậy GTN của \(Q=\dfrac{2011}{2a^2+2b^2+2008}\)là \(\dfrac{2011}{2012}\)khi a=b=1

Help me , please !!!!!!!!!!

1/\(=4a^2+4b^2+c^2+8ab-4bc-4ca+4b^2+4c^2+a^2+8bc-4ca-4ab+4a^2+4c^2+b^2+8ca-4bc-4ab=\)

\(=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)\)

2/

Ta có

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge0\)

\(\Leftrightarrow a^2+b^2+c^2\ge-2\left(ab+bc+ca\right)=2\)

\(\Rightarrow P=9\left(a^2+b^2+c^2\right)\ge18\)

\(\Rightarrow P_{min}=18\)

Ta có \(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

\(\Leftrightarrow\frac{x^2}{a^2+b^2+c^2}+\frac{y^2}{a^2+b^2+c^2}+\frac{z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

\(\Leftrightarrow\frac{x^2}{a^2+b^2+c^2}-\frac{x^2}{a^2}+\frac{y^2}{a^2+b^2+c^2}-\frac{y^2}{b^2}+\frac{z^2}{a^2+b^2+c^2}-\frac{z^2}{c^2}=0\)

\(\Leftrightarrow x^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\right)+y^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\right)+z^2\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\right)=0\)

Do \(\left\{\begin{matrix}\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\\\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\\\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\end{matrix}\right.\ne0\) và \(a,b,c\ne0\)

\(\Rightarrow\left\{\begin{matrix}x^2=0\\y^2=0\\z^2=0\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)

Ta có \(A=x^{2008}+y^{2008}+z^{2008}\)

\(\Rightarrow A=0+0+0\)

\(\Rightarrow A=0\)

Vậy A = 0