a, Ta có : \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=\left(x+y\right)\left(\left(x+y\right)^2-2xy-xy\right)\)

\(=1\left(1^2-3\left(-1\right)\right)=1\left(1^2+3\right)=4\)

b, Ta có : \(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\)

\(=\left(x-y\right)\left(\left(x-y\right)^2+3xy\right)\)

\(=1\left(1+3.9\right)=19\)

\(VT=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\\ =x^4-x^3y+x^3y-x^2y^2+x^2y^2-y^4\\ =\left(x^4-y^4\right)+\left(-x^3y+x^3y\right)+\left(-x^2y^2+x^2y^2\right)\\ =x^4-y^4=VP\)

\(VT=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\)

\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4\)

\(=x^4+\left(x^3y-x^3y\right)+\left(x^2y^2-x^2y^2\right)+\left(xy^3-xy^3\right)-y^4\)

\(=x^4+0+0+0-y^4\)

\(=x^4-y^4=VP\left(dpcm\right)\)

\(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)=x^4-y^4\\ \Leftrightarrow x.x^3+x.x^2y+x.xy^2+x.y^3-y.x^3-y.x^2y-y.xy^2-y.y^3=x^4-y^4\\ \Leftrightarrow x^4+x^3y+x^2y^2+xy^3-xy^3-x^2y^2-xy^3-y^4=x^4-y^4\\ \Leftrightarrow\left(x^4-y^4\right)+\left(x^3y-x^3y\right)+\left(x^2y^2-x^2y^2\right)+\left(xy^3-xy^3\right)=x^4-y^4\\ \Leftrightarrow x^4-y^4+0+0+0=x^4-y^4 \\ \Leftrightarrow x^4-y^4=x^4-y^4\left(đpcm\right)\)

Ta có \(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2=x+y+z\left(1\right)\)

Áp dụng bất đẳng thức buniacoxki ta có

\(\left(x^3+y^3+z^3\right)\left(x+y+z\right)\ge\left(x^2+y^2+z^2\right)^2\)

Kết hợp với (1)=> \(x^3+y^3+z^3\ge x^2+y^2+z^2\left(2\right)\)

\(\left(x^4+y^4+z^4\right)\left(x^2+y^2+z^2\right)\ge\left(a^3+b^3+c^3\right)^2\)

Kết hợp với (2)=> \(x^4+y^4+z^4\ge x^3+y^3+z^3\)(ĐPCM)

Dấu bằng xảy ra khi x=y=z=1

tick tui ,tui tick lai cho

Ta có:

\(x^4+y^4\ge x^3y+xy^3\Rightarrow2\left(x^4+y^4\right)\ge x^4+y^4+x^3y+xy^3=\left(x^3+y^3\right)\left(x+y\right)\)

\(\Rightarrow\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)

Σ\(\frac{x^4+y^4}{x^3+y^3}\)\(\ge x+y+z=2008\)

Hình như đề sai rồi

đúng đề mà bạn

Vậy b nói xem thử khi nào nó = 4.