a/ Ta có :

\(\left(x+y+t\right)-x^3-y^3-z^3=2011\)

\(\Leftrightarrow3\left(x+y\right)\left(y+t\right)\left(t+x\right)=2011\)

\(\Leftrightarrow\left(x+y\right)\left(y+t\right)\left(t+x\right)=\dfrac{2011}{3}\)

Thay vào D ta được :

\(D=\dfrac{2011}{\dfrac{2011}{3}}=3\)

Vậy.....

b/ Ta có :

\(H=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(\Leftrightarrow10899H=10899\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Leftrightarrow10899H=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)

\(\Leftrightarrow10899H=1+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+1+\dfrac{b}{c}+\dfrac{c}{b}+1\)

\(\Leftrightarrow10899H=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)

Áp dụng BĐT Cô - si cho các số dương ta có ;

\(+,\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)

+, \(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}=2\)

+, \(\dfrac{c}{a}+\dfrac{a}{c}\ge2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}=2\)

Cộng vế với vế của các BĐT ta có :

\(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}\ge6\)

\(\Leftrightarrow10899H\ge9\)

\(\Leftrightarrow H\ge\dfrac{1}{2011}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=6033\)

Vậy..

b ) Do a ; b ; c dương \(\Rightarrow\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\) dương

Áp dụng BĐT Cô - si cho 3 số dương , ta có :

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)

Theo GT : \(a+b+c=18099\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{18099}=\dfrac{1}{2011}\)

\(\Rightarrow H\ge\dfrac{1}{2011}\)

Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a+b+c=18099\\a=b=c\end{matrix}\right.\)

\(\Leftrightarrow a=b=c=6033\)

Vậy ...

Nhân 2 vế với \(\left(x-\sqrt{2011+x^2}\right)\) ta được:

\(\left(x^2-2011-x^2\right)\left(y+\sqrt{2011+y^2}\right)=2001\left(x-\sqrt{2011+x^2}\right)\)

\(\Leftrightarrow-2011\left(y+\sqrt{2011+y^2}\right)=2011\left(x-\sqrt{2011+x^2}\right)\)

\(\Leftrightarrow y+\sqrt{2011+y^2}=\sqrt{2011+x^2}-x\)(1)

Tương tự nhân 2 vế với \(\left(y-\sqrt{2011+y^2}\right)\) ta được:

\(x+\sqrt{2011+x^2}=\sqrt{2011+y^2}-y\)(2)

Cộng (1) và (2) vế theo vế ta được:

\(x+y=-x-y\)

\(\Leftrightarrow2\left(x+y\right)=0\)

\(\Leftrightarrow x+y=0\)

\(\Leftrightarrow x=-y\)

\(\Rightarrow T=-y^{2011}+y^{2011}=0\)

\(x^3+y^3+3xy\left(x+y\right)+\dfrac{1}{27}-3xy\left(x+y\right)-xy=0\)

\(\Leftrightarrow\left(x+y\right)^3+\dfrac{1}{27}-3xy\left(x+y+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left(x+y+\dfrac{1}{3}\right)\left[\left(x+y\right)^2-\dfrac{1}{3}\left(x+y\right)+\dfrac{1}{9}\right]-3xy\left(x+y+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow x^2+y^2-xy-\dfrac{1}{3}\left(x+y\right)+\dfrac{1}{9}=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-\dfrac{1}{3}\right)^2+\left(y-\dfrac{1}{3}\right)^2=0\)

\(\Leftrightarrow x=y=\dfrac{1}{3}\Rightarrow P=...\)

Từ \(x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{z}+\dfrac{1}{x}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=-2\) ta có:

\(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+2xyz=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

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

Không mất tính tổng quát, giả sử x + y = 0

\(\Leftrightarrow x=-y\)

\(\Leftrightarrow x^3=-y^3\).

Kết hợp với \(x^3+y^3+z^3=1\) ta có \(z^3=1\Leftrightarrow z=1\).

Vậy \(P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{-y}+\dfrac{1}{y}+\dfrac{1}{1}=1\).