Ta có :
\(\left(x+y\right)^5-x^5-y^5=0\)
\(\Leftrightarrow\left(x+y\right)^5-\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left[\left(x+y\right)^4-x^4+x^3y-x^2y^2+xy^3-y^4\right]=0\)
\(\Leftrightarrow\left(x+y\right)\left(5x^3+5x^2y^2+5xy^3\right)=0\)
\(\Leftrightarrow5xy\left(x+y\right)\left(x^2+xy+y^2\right)=0\left(1\right)\)
\(x,y\ne0,x^2+xy+y^2=\left(x+y\right)^2+\dfrac{3y^2}{4}\ne0\)
Nên \(\left(1\right)=>x+y=0\)
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\((x+y)^5-x^5-y^5=0 \\\Leftrightarrow x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5-x^5-y^5=0 \\\Leftrightarrow (5x^4y+5xy^4)+(10x^3y^2+10x^2y^3)=0 \\\Leftrightarrow 5xy(x^3+y^3)+10x^2y^2(x+y)=0 \\\Leftrightarrow 5xy(x+y)(x^2-xy+y^2)+10x^2y^2(x+y)=0 \\\Leftrightarrow 5xy(x+y)(x^2-xy+y^2+2xy)=0 \\\Leftrightarrow 5xy(x+y)(x^2+xy+y^2)=0 \\\Leftrightarrow \left[\begin{matrix}5xy=0\Rightarrow x=0 \ or \ y=0\\ x+y=0\\ x^2+xy+y^2=0\end{matrix}\right.\)
Mà \(x,y\neq 0\)\(x^2+xy+y^2=x^2+2.x.\dfrac{y}{2}+\dfrac{y^2}{4}+\dfrac{3y^2}{4}=(x+\dfrac{y}{2})^2+\dfrac{3y^2}{4}>0\)
\(\Rightarrow x+y=0\)
