2) \(\hept{\begin{cases}^{x^2-xy=y^2-yz}\left(1\right)\\^{y^2-yz=z^2-zx}\left(2\right)\\^{z^2-zx=x^2-xy}\left(3\right)\end{cases}}\)
lấy (2) - (1) suy ra\(2yz=2y^2+xy+xz-x^2-z^2\)
lấy (3) - (1) suy ra \(2xy=zx+yz-z^2+2x^2-y^2\)
lấy (3) - (2) suy ra \(2zx=xy+yz+2z^2-x^2-y^2\)
cộng lại đc \(yz+xz+xy=0\) do đó \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{yz+xz+xy}{xyz}=0\)
1) \(a=x^2-xy=x\left(x-y\right)\ne0\left(x\ne0,x\ne y\right)\)
mik cần c3 , ai làm giúp mik đc ko
Từ xy + yz + zx = 0 => xz = - ( yz + xy )
Có \(\frac{1}{x}\)+ \(\frac{1}{y}\) + \(\frac{1}{z}\) = 0 => \(\frac{1}{x}\) = - \(\frac{1}{z}\) - \(\frac{1}{y}\)
=> \(\frac{y}{x}\)= - 1 - \(\frac{y}{z}\)
CMTT : \(\frac{x}{z}\) = - 1 - \(\frac{x}{y}\)
\(\frac{z}{y}\)= -1 - \(\frac{z}{x}\)
=> M = \(-3\) \(-\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)\)
Đặt A = \(\frac{x}{y}\) + \(\frac{y}{z}\) + \(\frac{z}{x}\)
A = \(\frac{\text{x^2z + y^2x + z^2y }}{xyz}\)
A = \(\frac{-x\left(xy+yz\right)+xy^2+yz^2}{xyz}\)
A = \(\frac{-\left(x^2y+xyz\right)+xy^2+yz^2}{xyz}\)
A = \(\frac{-x^2-xz+xy+z^2}{xz}\) ( Rút gọn)
A = \(\frac{a-a}{xz}\) = 0
=> M = - 3 - A = -3 - 0 = -3
Vậy M = -3
