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a)
Có \(x+y+z=0\)
\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz=0\)
\(\Rightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\) (1)
Phân tích :
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\)
\(=x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2\)
\(=2\left(x^2+y^2+z^2\right)+\left[-2\left(xy+yz+xz\right)\right]\)(Áp dung (1)ta được :)
\(=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2\)
\(=3\left(x^2+y^2+z^2\right)\)
\(\Rightarrow P=\dfrac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
\(\Rightarrow P=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}\)
\(\Rightarrow P=\dfrac{1}{3}\)
\(x+y+z=0\)
\(\Rightarrow\left(x+y+z\right)^2=0\)
\(x^2+y^2+z^2+2\left(xy+yz+zx\right)=0\)
\(x^2+y^2+z^2=-2\left(xy+yz+zx\right)\)
\(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
\(=\frac{-2\left(xy+yz+zx\right)}{2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)}\)
\(=\frac{-2\left(xy+yz+zx\right)}{2\left[-2\left(xy+yz+zx\right)\right]-2\left(xy+yz+xz\right)}\)
\(=\frac{-2\left(xy+yz+zx\right)}{-4\left(xy+yz+zx\right)-2\left(xy+yz+xz\right)}\)
\(=\frac{-2\left(xy+yz+zx\right)}{-6\left(xy+yz+zx\right)}\)
\(=\frac{1}{3}\)
Ta có: \(x+y+z=0\)
\(\Rightarrow x+y=-z\)
\(\Rightarrow\left(x+y\right)^2=\left(-z\right)^2\)
\(x^2+2xy+y^2=z^2\)
\(x^2+y^2-z^2=-2xy\)
\(\frac{2x^2y+2xy^2}{x^2+y^2-z^2}\)
\(=\frac{2xy\left(x+y\right)}{-2xy}\)
\(=\frac{-2xyz}{-2xy}\)
\(=z\)
Cho x+y+z=0 Rút gọn:\(\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
Ta có: \(x+y+z=0\Rightarrow\hept{\begin{cases}-x=y-z\\-y=z-x\\-z=x-y\end{cases}}\)
Mà \(x^2=\left(-x\right)^2;y^2=\left(-y\right)^2;z^2=\left(-z\right)^2\)
Thế vào biểu thức, ta được:
\(\frac{x^2+y^2+z^2}{x^2+y^2+z^2}=1\)
\(x+y+z=0\Rightarrow\hept{\begin{cases}x=-\left(y+z\right)\\y=-\left(z+x\right)\\z=-\left(x+y\right)\end{cases}}\)
\(\Rightarrow P=\frac{x^2+y^2+z^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{\left[-\left(y+z\right)\right]^2+\left[-\left(z+x\right)\right]^2+\left[-\left(x+y\right)\right]^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{\left(y+z\right)^2+\left(z+x\right)^2\left(x+y\right)^2}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{-\left[\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\right]}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=-1\)
Mik mới biết làm câu a thôi còn câu b thì từ từ mik nghĩ đã nhé @-@
Chúc bn học giỏi nhoa!!!