b) \(x^3+x^2y-2x^2-xy-y^2+3y+x-1\)
\(=x^2\left(x+y-2\right)-xy-y^2+3y+x-1\)
\(=-xy-y^2+3y+x-1\)
\(=-\left(xy+y^2-3y-x+1\right)\)
\(=-\left[y\left(x+y-2\right)-y-x+1\right]\)
\(=x+y-1=x+y-2+1=0+1=1\)
Vậy giá trị đa thức luôn là hằng số
a) Ta có:
\(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=b=c\\a+b+c=0\end{matrix}\right.\)(đpcm)
a,
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2=ab+bc+ca\end{matrix}\right.\)
Có \(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ca\end{matrix}\right.\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
Dấu "=" khi a=b=c
Thay lại ta có điều phải chứng minh
b, \(x^3+x^2y-2x^2-xy-y^2+3y+x-1\)
\(=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-2\right)+1=1\)
