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Ta có:
\(M=\frac{x\left(yz-x^2\right)+y\left(zx-y^2\right)+z\left(xy-z^2\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{xyz-x^3+xyz-y^3+xyz-z^3}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{3xyz-x^3-y^3-z^3}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
\(-M=\frac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
Xét đẳng thức phụ:
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=\left[\left(a +b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2-ab\right]=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(=\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-abc-ac\right)\)
\(=\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)
Thay vào -M ta có:
\(-M=\frac{\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{1}{2}\left(x+y+z\right)\Rightarrow M=-\frac{1}{2}\left(x+y+z\right)\)
Giờ thay: \(x=2014^{2015}-20142015;y=20142015-2015^{2014};z=2015^{2014}-2014^{2015}\)
Ta có:
\(M=-\frac{1}{2}\left(2014^{2015}-20142015+20142015-2015^{2014}+2015^{2014}-2014^{2015}\right)=0\)
Ta có :
\(VT=\left(x+y\right)\left(y+z\right)\left(z+x\right)+xyz\)
\(=\left(xy+y^2+xz+yz\right)\left(z+x\right)+xyz\)
\(=xyz+y^2z+xz^2+yz^2+x^2y+y^2x+x^2z+xyz+xyz\)
\(=\left(x^2y+xyz+x^2z\right)+\left(y^2x+y^2z+xyz\right)+\left(xyz+z^2y+z^2x\right)\)\(=x\left(xy+yz+zx\right)+y\left(xy+yz+zx\right)+z\left(xy+yz+zx\right)\)
\(=\left(x+y+z\right)\left(xy+yz+zx\right)=VP\)
\(\left(đpcm\right)\)
:D
Đẳng thức ban đầu \(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=4x^2+4y^2+4z^2-4xy-4yz-4zx\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\)
\(\Leftrightarrow x=y=z\)
Ta có:
\(\left(x+y+z\right)\left(xy+yz+zx\right)=xyz\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow x=-y;y=-z;z=-x\)
Với \(x=-y\)
\(\Rightarrow x^{2017}+y^{2017}+z^{2017}=z^{2017}=\left(x+y+z\right)^{2017}\)
Tương tự cho 2 trường hợp còn lại
Lời giải:
Ta có:
\(x^2+2y^2+z^2-2xy-2y-4z+5=0\)
\(\Leftrightarrow (x^2+y^2-2xy)+(y^2-2y+1)+(z^2-4z+4)=0\)
\(\Leftrightarrow (x-y)^2+(y-1)^2+(z-2)^2=0\)
Ta thấy:
\(\left\{\begin{matrix} (x-y)^2\geq 0\\ (y-1)^2\geq 0\\ (z-2)^2\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{R}\)
\(\Rightarrow (x-y)^2+(y-1)^2+(z-2)^2\geq 0\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} x-y=0\\ y-1=0\\ z-2=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=1\\ y=1\\ z=2\end{matrix}\right.\)
Do đó:
\(A=(x-1)^{2015}+(y-1)^{2015}+(z-1)^{2015}=1\)
ta có (x+y+z).(xy+yz+zx) - xyz = 0
<=> (x+y).(y+z).(z+x) = 0
=> vế trái phải có 1 nhân tử bằng 0 ,chẳng hạn x + y = 0 => x = -y
=> x^2013 = -y^2013
=> x^2013 + y^2013 + z^2013 = - y^2013 + y^2013 + z^2013 + = z^2013 = ( x +y + z )^2013
a) \(\left(x-y\right)^2+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-y\right)^2+\left(y-z+z-x\right)\left[\left(y-z\right)^2-\left(y-z\right)\left(z-x\right)+\left(z-x\right)^2\right]\)
\(=\left(x-y\right)^2+\left(y-x\right)\left(x^2+y^2+3z^2-3yz+xy-3xz\right)\)
\(=\left(x-y\right)\left(x-y-x^2-y^2-3z^2+3yz-xy+3xz\right)\)
Cô nghĩ phân tích đa thức này sẽ đẹp hơn:
\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3\)
\(=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3\)
\(=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(z-x\right)^2\right]\)
\(=\left(x-z\right)\left(3y^2-3xy+3zx-3xyz\right)\)
\(=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
b) \(\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)
\(=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz\)
\(=xy\left(x+y+z\right)+\left(yz+zx\right)\left(x+y+z\right)-xyz\)
\(=xy\left(x+y+z-z\right)+\left(yz+zx\right)\left(x+y+z\right)\)
\(=xy\left(x+y\right)+z\left(y+x\right)\left(x+y+z\right)\)
\(=\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=\left(x+y\right)\left(xy+zx+zy+z^2\right)\)
\(=\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
a) \left(x-y\right)^2+\left(y-z\right)^3+\left(z-x\right)^3(x−y)2+(y−z)3+(z−x)3
=\left(x-y\right)^2+\left(y-z+z-x\right)\left[\left(y-z\right)^2-\left(y-z\right)\left(z-x\right)+\left(z-x\right)^2\right]=(x−y)2+(y−z+z−x)[(y−z)2−(y−z)(z−x)+(z−x)2]
=\left(x-y\right)^2+\left(y-x\right)\left(x^2+y^2+3z^2-3yz+xy-3xz\right)=(x−y)2+(y−x)(x2+y2+3z2−3yz+xy−3xz)
=\left(x-y\right)\left(x-y-x^2-y^2-3z^2+3yz-xy+3xz\right)=(x−y)(x−y−x2−y2−3z2+3yz−xy+3xz
\left(x-y\right)^3+\left(y-z\right)^3+\left
=\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3
=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\l
=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(z-x\
=\left(x-z\right)\left(
=3\left(x-y\right)\lefb) \left(x+y+z\right)\left(xy+yz+zx\right)-xyzb)(x+y+z)(xy+yz+zx)−xyz
=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz=(xy+yz+zx)(x+y+z)−xyz
=xy\left(x+y+z\right)+\left(yz+zx\right)\left(x+y+z\right)-xyz=xy(x+y+z)+(yz+zx)(x+y+z)−xyz
=xy\left(x+y+z-z\right)+\left(yz+zx\right)\left(x+y+z\right)=xy(x+y+z−z)+(yz+zx)(x+y+z)
=xy\left(x+y\right)+z\left(y+x\right)\left(x+y+z\right)=xy(x+y)+z(y+x)(x+y+z)
=\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]=(x+y)[xy+z(x+y+z)]
=\left(x+y\right)\left(xy+zx+zy+z^2\right)=(x+y)(xy+zx+zy+z2)
=\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]=(x+y)[x(y+z)+z(y+z)]
=\left(x+y\right)\left(y+z\right)\left(z+x\right)=(x+y)(y+z)(z+x)