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2 tháng 7 2021
a) xy(x + y) + yz(y + z) + xz(z + x) + 3xyz
= xy(X + y + z) + yz(x + y + z) + xz(X + y + z)
= (x + y +z)(xy + yz+ xz)
b) xy(x + y) - yz(y + z) - xz(z - x)
= x2y + xy2 - y2z - yz2 - xz2 + x2z
= x2(y + z) - yz(y + z) + x(y2 - z2)
= x2(y + z) - yz(y + z) + x(y + z)(y - z)
= (y + z)(x2 - yz + xy - xz)
= (y + z)[x(x + y) - z(x + y)]
= (y + z)(x + y)(x - z)
c) x(y2 - z2) + y(z2 - x2) + z(x2 - y2)
= x(y - z)(y + z) + yz2 - yx2 + x2z - y2z
= x(y - z)(y + z) - yz(y - z) - x2(y - z)
= (y - z)((xy + xz - yz - x2)
= (y - z)[x(y - x) - z(y - x)]
= (y - z)(x - z)(y -x)
5 tháng 11 2016
\(\frac{x^2-yz}{yz}+1+\frac{y^2-zx}{zx}+1+\frac{z^2-xy}{xy}+1=3\Leftrightarrow\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3\)
\(\Leftrightarrow\frac{1}{xyz}\left(x^3+y^3+z^3\right)=3\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
Tới đây bạn thay vào nhé :)
2 tháng 8 2019
Đẳng thức ban đầu \(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=4x^2+4y^2+4z^2-4xy-4yz-4zx\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\)
\(\Leftrightarrow x=y=z\)
KK
24 tháng 1 2017
\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)
\(\Leftrightarrow\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x^2-yz}{x-xyz}=\frac{y^2-xz}{y-xyz}=\frac{x^2-y^2+xz-yz}{x-xyz-y+xyz}=\frac{\left(x-y\right)\left(x+y\right)+z\left(x-y\right)}{x-y}=\frac{\left(x-y\right)\left(x+y+z\right)}{x-y}=x+y+z\)
\(\Rightarrow\frac{x^2-yz}{x-xyz}=x+y+z\)
\(\Rightarrow x^2-yz=\left(x-xyz\right)\left(x+y+z\right)\)
\(\Rightarrow x^2-yz=x\left(x-xyz\right)+y\left(x-xyz\right)+z\left(x-xyz\right)\)
\(\Rightarrow x^2-yz=x^2-x^2yz+xy-xy^2z+xz-xyz^2\)
\(\Rightarrow-yz-xy-xz=-x^2yz-xy^2z-xyz^2\)
\(\Rightarrow-\left(yz+xy+xz\right)=-\left(x^2yz+xy^2z+xyz^2\right)\)
\(\Rightarrow yz+xy+xz=x^2yz+xy^2z+xyz^2\)
\(\Rightarrow yz+xy+xz=xyz\left(x+y+z\right)\)
Vậy nếu \(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\) thì \(yz+xy+xz=xyz\left(x+y+z\right)\)
2 tháng 9 2018
\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
CH
Cô Hoàng Huyền
Admin
VIP
26 tháng 9 2017
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) \(\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)
\(=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz\)
\(=xy\left(x+y+z\right)+\left(yz+zx\right)\left(x+y+z\right)-xyz\)
\(=xy\left(x+y+z-z\right)+\left(yz+zx\right)\left(x+y+z\right)\)
\(=xy\left(x+y\right)+z\left(y+x\right)\left(x+y+z\right)\)
\(=\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=\left(x+y\right)\left(xy+zx+zy+z^2\right)\)
\(=\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
6 tháng 7 2022
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)
M
15 tháng 8 2018
Ta có : \(\left(x+y+z\right)^2=x^2+y^2+z^2.\)
<=> \(x^2+2xy+y^2+2xz+2yz+z^2-x^2-y^2-z^2=0\)
<=> \(2xy+2xz+2yz=0\)
<=> \(2.\left(xy+xz+yz\right)=0\)
<=> \(xy+xz+yz=0\)
Vậy_
BD
15 tháng 8 2018
Ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2\)
\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+xz+yz\right)=x^2+y^2+z^2\)
\(\Leftrightarrow2\left(xy+xz+yz\right)=0\)
\(xy+xz+yz=0\left(đpcm\right)\)
Ta có: \(\left(x+y+z\right)^2=3\left(xy+yz+xz\right)\)
\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=3\left(xy+yz+xz\right)\)
\(\Leftrightarrow x^2+y^2+z^2=xy+yz+xz\)
\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2xz\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)
\(\Leftrightarrow x^2+x^2+y^2+y^2+z^2+z^2-2xy-2xz-2yz=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
Vì \(\left(x-y\right)^2;\left(y-z\right)^2\)và \(\left(x-z\right)^2\)luôn luôn \(\ge0\)
để phương trình trên bằng 0 thì \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(x-z\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}\Rightarrow}\hept{\begin{cases}x=y\\y=z\\x=z\end{cases}}}\)
\(\Rightarrow x=y=z\left(ĐPCM\right)\)