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![](https://rs.olm.vn/images/avt/0.png?1311)
Xét \(x+y+z=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+z=-x\\z+x=-y\\x+y=-z\end{matrix}\right.\)
\(\Rightarrow A=\left(2-1\right)\left(2-1\right)\left(2-1\right)=1\)
Xét \(x+y+z\ne0\) thì ta có:
\(\dfrac{x}{y+z+3x}=\dfrac{y}{z+x+3y}=\dfrac{z}{x+y+3z}=\dfrac{x+y+z}{5x+5y+5z}=\dfrac{x+y+z}{5\left(x+y+z\right)}=\dfrac{1}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}5x=y+z+3x\\5y=z+x+3y\\5z=x+y+3z\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=y+z\\2y=z+x\\2z=x+y\end{matrix}\right.\)
\(\Rightarrow A=\left(2+2\right)\left(2+2\right)\left(2+2\right)=64\)
Vậy \(\left[{}\begin{matrix}A=1\\A=64\end{matrix}\right.\)
Nếu bị lỗi thì bạn có thể xem đây nhé:
![](https://rs.olm.vn/images/avt/0.png?1311)
x-y-z=0
\(\Rightarrow x=y+z\)
\(\Rightarrow y=x-z\)
\(\Rightarrow-z=y-z\)
\(B=\left(1-\dfrac{z}{x}\right).\left(1-\dfrac{y}{x}\right).\left(1+\dfrac{y}{z}\right)\)
\(B=\left(\dfrac{x-z}{x}\right).\left(\dfrac{y-x}{y}\right).\left(\dfrac{z+y}{z}\right)\)
\(B=(\dfrac{y}{x}).\left(\dfrac{-z}{y}\right).\left(\dfrac{x}{z}\right)\)
\(B=\dfrac{\left(y.x.-z\right)}{\left(y.x.z\right)}\Rightarrow B=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\left(\dfrac{x+2y}{y}\right)\left(\dfrac{y+2z}{z}\right)\left(\dfrac{z+2x}{x}\right)\)
Ta có
\(\dfrac{x+2y-z}{z}=\dfrac{y+2z-x}{x}=\dfrac{z+2x-y}{y}=\)
\(=\dfrac{x+2y-z+y+2z-x+z+2x-y}{x+y+z}=\)
\(=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
\(\Rightarrow\dfrac{x+2y}{z}-1=\dfrac{y+2x}{x}-1=\dfrac{z+2x}{y}-1=2\)
\(\Rightarrow\dfrac{x+2y}{z}=\dfrac{y+2x}{x}=\dfrac{z+2x}{y}=3\)
\(\Rightarrow P=3.3.3=27\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : từ x - y - z =0
\(\Rightarrow x-z=y\) ; \(-z=y-x\) ; \(y+z=x\)
Lại có \(B=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(\Rightarrow B=\dfrac{x-z}{x}.\dfrac{y-x}{y}.\dfrac{y+z}{z}\)
thay các hằng đẳng thức vừa tìm được vào B
\(\Rightarrow B=\dfrac{y}{x}.\dfrac{-z}{y}.\dfrac{x}{z}=-1\)
vậy B = -1
tik mik nha !!!
Với \(x,y,z\ne0\), ta có: \(x-y-z=0\Leftrightarrow\left\{{}\begin{matrix}x-z=y\\y-x=-z\\z+y=x\end{matrix}\right.\)\((*)\)
Mặt khác: \(B=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(=\dfrac{x-z}{x}\cdot\dfrac{y-x}{y}\cdot\dfrac{z+y}{z}\)
Thay \((*)\) vào \(B\), ta được:
\(B=\dfrac{y}{x}\cdot\dfrac{-z}{y}\cdot\dfrac{x}{z}=-1\)
Vậy \(B=-1\) thoả mãn đề bài.
+ \(x-y-z=0\)
\(\Rightarrow\dfrac{x}{x}-\dfrac{\left(y+z\right)}{x}=0\) (Do \(x\ne0\))
\(\Leftrightarrow1-\dfrac{y+z}{x}=0\)
+ \(x-y-z=0\)
\(\Rightarrow\dfrac{\left(x-z\right)}{y}-\dfrac{y}{y}=0\) (Do \(y\ne0\))
\(\Leftrightarrow1-\dfrac{x-z}{y}=0\)
+ \(x-y-z=0\)
\(\Rightarrow\dfrac{\left(x-y\right)}{z}-\dfrac{z}{z}=0\) (Do \(z\ne0\))
\(\Leftrightarrow1-\dfrac{x-y}{z}=0\)
Ta có: \(B=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(=\left(1-\dfrac{x}{y}-\dfrac{z}{x}+\dfrac{zx}{xy}\right)\left(1+\dfrac{y}{z}\right)\)
\(=\left(1-\dfrac{x}{y}-\dfrac{z}{x}+\dfrac{z}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(=1+\dfrac{y}{z}-\dfrac{x}{y}-\dfrac{xy}{yz}-\dfrac{z}{x}-\dfrac{zy}{xz}+\dfrac{z}{y}+\dfrac{zy}{yz}\)
\(=1-\dfrac{y+z}{x}+1-\dfrac{x-z}{y}+1-\dfrac{x-y}{z}-1\)
\(=-1\)
Vậy \(B=-1\)