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A = (1−zx )(1xy )(1+yz )
A = \(\frac{x-z}{x}\). \(\frac{x}{y}\). \(\frac{z+y}{z}\)
Mà x-y-z = 0
=> x-z = y ; z+y=x
Ta có A= \(\frac{y}{x}\). \(\frac{x}{y}\). \(\frac{x}{z}\)= 1.\(\frac{x}{z}\)= \(\frac{x}{z}\)
#)Giải :
\(A=\left(1-\frac{z}{y}\right).\left(1-\frac{x}{y}\right).\left(1-\frac{y}{z}\right)\)
\(A=\frac{x-z}{x}.\frac{x+y}{z}.\frac{z-y}{x}\)
\(x+y-z=0\Leftrightarrow\hept{\begin{cases}x+y=z\\x-z=-y\\z-y=x\end{cases}}\)
Thay vào A, ta được :
\(A=\frac{-y}{x}.\frac{z}{y}.\frac{x}{z}=\frac{-yzx}{xyz}=-1\)
~Will~be~Pens~
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{x+y-2014z}{z}=\frac{y+z-2014x}{x}=\frac{z+x-2014y}{y}=\frac{\left(-2012\right)\left(x+y+z\right)}{x+y+z}=-2012\)
Ta có: \(\frac{x+y-2014z}{z}=-2012\Rightarrow x+y-2014z=-2012z\Leftrightarrow x+y=2z\)
Tương tự: \(y+z=2x,z+x=2y\)
Khi đó: \(A=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}=\frac{2x.2y.2z}{xyz}=8\)
Vậy A=8.
Nguyễn Tất Đạt thiếu 1 trường hợp nha bạn
\(x+y+z=0\)
\(\Rightarrow\hept{\begin{cases}x=-y-z\\y=-x-z\\z=-x-y\end{cases}}\)
\(A=\left(1+\frac{-y-z}{y}\right).\left(1+\frac{-x-z}{z}\right).\left(1+\frac{-x-y}{x}\right)\)
\(A=\left(-\frac{z}{y}\right).\left(\frac{-x}{z}\right).\left(\frac{-y}{x}\right)=-1\)
\(\text{Ta có: }x-y-z=0\Rightarrow x=y+z\)
\(y=x-z\)
\(z=x-y\)
\(\text{Mặt khác: }A=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)
\(=\left(\frac{x}{x}-\frac{z}{x}\right)\left(\frac{y}{y}-\frac{x}{y}\right)\left(\frac{z}{z}+\frac{y}{z}\right)\)
\(=\frac{x-z}{x}.\frac{y-x}{y}.\frac{y+z}{z}\)
\(=\frac{x-z}{y+z}.\frac{y-x}{x-z}.\frac{y+z}{x-y}\)
\(=\frac{x-z}{y+z}.\frac{y-x}{x-z}.\frac{y+z}{-\left(y-x\right)}\)
\(=-1\)
Ta có: x - y - z = 0 \(\Rightarrow\begin{cases}x-z=y\\y-x=-z\\z+y=x\end{cases}\)
\(A=\left(1-\frac{z}{x}\right).\left(1-\frac{x}{y}\right).\left(1+\frac{y}{z}\right)\)
\(A=\frac{x-z}{x}.\frac{y-x}{y}.\frac{z+y}{z}\)
\(A=\frac{y}{x}.\frac{-z}{y}.\frac{x}{z}=-1\)
Ta có \(x-y-z=0\)
\(\Rightarrow\hept{\begin{cases}x-z=y\\y-x=-z\\z+y=x\end{cases}}\)( 1 )
Ta có:
\(B=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)
\(B=\frac{x-z}{x}.\frac{y-x}{y}.\frac{z+y}{z}\)
Thay điều ( 1 ) vào biểu thức ta có:
\(B=\frac{x-z}{x}.\frac{y-x}{y}.\frac{z+y}{z}\)
\(\Rightarrow B=\frac{y}{x}.\frac{-z}{y}.\frac{x}{z}\)
\(\Rightarrow B=-1\)
Vậy B = -1
\(x-y-z=0\Rightarrow y-z=x;x-z=y;x=y+z;y-x=-z\)
\(B=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)=\frac{x-z}{x}.\frac{y-x}{y}.\frac{y+z}{z}=\frac{y}{x}.\frac{-z}{y}.\frac{x}{z}=-1\)
Vậy B = - 1
A=(x-z/x).(y-x/y)(z+y/z)
ta có x-y-z=0
=>x-z=y
=>y-x=-z
=>z+y=x
=>A=y/x.-z/y.x/z=-1