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x+y+z=0
nên x+y=-z; y+z=-x; x+z=-y
\(\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)
\(=\dfrac{x+y}{y}\cdot\dfrac{y+z}{z}\cdot\dfrac{x+z}{x}=-1\)
Từ \(x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{z}+\dfrac{1}{x}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=-2\) ta có:
\(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+2xyz=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\).
Không mất tính tổng quát, giả sử x + y = 0
\(\Leftrightarrow x=-y\)
\(\Leftrightarrow x^3=-y^3\).
Kết hợp với \(x^3+y^3+z^3=1\) ta có \(z^3=1\Leftrightarrow z=1\).
Vậy \(P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{-y}+\dfrac{1}{y}+\dfrac{1}{1}=1\).
b: \(M=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=\dfrac{a+b+c}{abc}=0\)
c: \(B=\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(x-z\right)\left(y-z\right)}-\dfrac{x}{\left(x-z\right)\left(x-y\right)}\)
\(=\dfrac{y\left(x-z\right)-z\left(x-y\right)-x\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+zy-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)
a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
=0
c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{1}{xyz}\)
`@ x+y+z=1`.
`<=>` \(\left\{{}\begin{matrix}x=1-y-z\\y=1-z-x\\z=1-x-y\end{matrix}\right.\)
`P=(x+y)^2/(xy+1-x-y).(y+z)^2/(yz-y-z+1).(x+z)^2/(xy-x-y+1)`.
`<=> ((1-z)^2(1-y)^2(1-x)^2)/((1-x)(1-y)(1-y)(1-z)(1-z)(1-x).`
`=1.`
Vậy `P` không phụ thuộc vào giá trị của biến.
* Nếu x + y + z = 0
\(A=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{z}{y}\right)\left(1+\dfrac{x}{z}\right)\)
\(=\dfrac{x+y}{x}\cdot\dfrac{y+z}{y}\cdot\dfrac{z+x}{z}=\dfrac{\left(-z\right)}{x}\cdot\dfrac{\left(-x\right)}{y}\cdot\dfrac{\left(-y\right)}{z}=\dfrac{\left(-x\right)\left(-y\right)\left(-z\right)}{xyz}=-\dfrac{xyz}{xyz}=-1\)
* Nếu x + y + z khác 0
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x-y-z}{x}=\dfrac{y-x-z}{y}=\dfrac{-x-y+z}{z}=\dfrac{x-y-z+y-x-z-x-y+z}{x+y+z}=\dfrac{-x-y-z}{x+y+z}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x-y-z=-x\\y-x-z=-y\\-x-y+z=-z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+z=2x\\z+x=2y\\x+y=2z\end{matrix}\right.\Rightarrow x=y=z\)
\(\Rightarrow A=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{z}{y}\right)\left(1+\dfrac{x}{z}\right)\)
\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2\cdot2\cdot2=8\)
a: \(=\dfrac{1}{\left(x-y\right)\left(y-z\right)}-\dfrac{1}{\left(y-z\right)\left(x-z\right)}-\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)
b: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(x-y\right)\left(y-z\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{z\left(y^2-x^2\right)-z^2\left(y-x\right)-xy\left(y-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{\left(x-y\right)\left[-z\left(x+y\right)+z^2+xy\right]}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{-zx-zy+z^2+xy}{xyz\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{z\left(z-x\right)-y\left(z-x\right)}{xyz\left(y-z\right)\left(x-z\right)}=\dfrac{1}{xyz}\)
\(P=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{x}{z}\right)\\ =\dfrac{x+y}{y}.\dfrac{y+z}{z}.\dfrac{x+z}{z}\\ =\dfrac{-z}{y}.\dfrac{-x}{z}.\dfrac{-y}{x}\\ =\dfrac{-xyz}{xyz}=-1\)
xuczat :) mk ko hiểu ạ