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tích mình đi

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mình tích lại 

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sao ko ai trả lời vậy

Từ giải thiết, ta suy ra được những điều sau :

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

\(=\frac{x}{\left[y-\left(x+y\right)\right]\left(y^2+y+1\right)}-\frac{y}{\left[x-\left(x+y\right)\right]\left(x^2+x+1\right)}\)

\(=\frac{x}{-x\left(y^2+y+1\right)}-\frac{y}{-y\left(x^2+x+1\right)}\)

\(=\frac{-1}{y^2+y+1}+\frac{1}{x^2+x+1}\)      (1)

Và \(\left(x^2+x+1\right)\left(y^2+y+1\right)\) 

\(=x^2y^2+x^2y+x^2+xy^2+xy+x+y^2+y+1\)

\(=x^2y^2+\left(x^2+xy\left(x+y\right)+xy+y^2\right)+\left(x+y\right)+1\)

\(=x^2y^2+\left(x^2+2xy+y^2\right)+1+1\)

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

\(=x^2y^2+3\)   (2)

Từ (1) và (2) suy ra :

\(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}\)

\(=\frac{-1}{y^2+y+1}+\frac{1}{x^2+x+1}+\frac{2\left(x-y\right)}{\left(y^2+y+1\right)\left(x^2+x+1\right)}\)

\(=\frac{-x^2-x-1+y^2+y+1+2x-2y}{\left(y^2+y+1\right)\left(x^2+x+1\right)}\)

\(=\frac{-x^2+y^2+x-y}{\left(y^2+y+1\right)\left(x^2+x+1\right)}\)

\(=\frac{\left(x+y\right)\left(y-x\right)+x-y}{\left(y^2+y+1\right)\left(x^2+x+1\right)}\)

\(=\frac{y-x+x-y}{\left(y^2+y+1\right)\left(x^2+x+1\right)}\)

\(=0\)(ĐPCM)

Biến đổi

\(\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{x^4-x-y^4+y}{\left(x^3-1\right)\left(y^3-1\right)}=\frac{\left(x^4-y^4\right)-\left(x-y\right)}{xy\left(y^2+y+1\right)\left(x^2+x+1\right)}\)

(do x+y=1 => y-1=-x và x-1=-y)

\(=\frac{\left(x-y\right)\left(x+y\right)\left(x^3+y^3\right)-\left(x-y\right)}{xy\left(x^2y^2+y^2x+y^2+yx^2+xy+y+x^2+x+1\right)}\)

\(=\frac{\left(x-y\right)\left(x^2+y^2-1\right)}{xy\left[x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+2\right]}\)

\(=\frac{\left(x-y\right)\left(x^2-x+y^2-y\right)}{xy\left[x^2y^2+\left(x+y\right)^2+2\right]}=\frac{\left(x-y\right)\left[x\left(x-1\right)+y\left(y-1\right)\right]}{xy\left(x^2y^2+3\right)}\)

\(=\frac{\left(x-y\right)\left[x\left(-y\right)+y\left(-x\right)\right]}{xy\left(x^2y^2+3\right)}=\frac{\left(x-y\right)\left(-2xy\right)}{xy\left(x^2y^2+1\right)}=\frac{-2\left(x-y\right)}{x^2y^2+3}\)

=> ĐPCM

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

Lại có \(\left(x^2+y^2\right)^2=\left(\sqrt{x}.\sqrt{x^3}+\sqrt{y}\sqrt{y^3}\right)^2\le\left(x+y\right)\left(x^3+y^3\right)\)

\(\Rightarrow\left(x^2+y^2\right)^2\le\left(x+y\right)\left(x^2+y^2\right)\Rightarrow x^2+y^2\le x+y\)

\(\Rightarrow\left(x^2+y^2\right)^2\le\left(x+y\right)^2\le2\left(x^2+y^2\right)\)

\(\Rightarrow x^2+y^2\le2\Rightarrow x^3+y^3\le2\)

Dấu "=" xảy ra khi \(x=y=1\)