\(\left(x+y+z\right)^2=x^2+y^2+z^2\\ \Rightarrow x^2+y^2+z^2+2xy+2xz+2yz=x^2+y^2+z^2\\ \Rightarrow2xy+2xz+2yz=0\\ \Rightarrow xy+xz+yz=0\left(đpcm\right)\)
Rút gọn M
M= \(\dfrac{x\left(yz-x^2\right)+y\left(zx-y^2\right)+z\left(xy-z^2\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
Chứng minh rằng:
Nếu \(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cx\right)^2\) thì \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\).
Chứng minh rằng:
a) x = y = z , biết :
x + y + z = 0 và xy + yz + zx = 0
b) (x+y)2 ≥ 4xy
Cho \(x^2+y^2+z^2=xy+yz+zx\). Chứng minh \(x=y=z\).
CMR: x2+y2+z2-xy-yz-xz=\(\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{2}\)
Chứng minh đẳng thức:
( x+y+z)2-x2-y2-z2= 2( xy+yz+zx)
cho x+y+z=0
và xy+yz+zx=0
Chứng minh x=y=z
2.CMR: Nếu x2+y2+z2 = xy + yz + zx thì x=y=z
Giup mk vs
Cho \(x,y,z\ne0;x\ne y.CMR\):
Nếu \(\dfrac{x^2-yz}{x\left(1-yz\right)}=\dfrac{y^2-xz}{y\left(1-xz\right)}\) thì \(x+y+z=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
