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Ta có: x-y-z=0 <=> x=y+z Thay vào A ta có:
A=\(\left(1-\dfrac{z}{y+z}\right)\left(1-\dfrac{y+z}{y}\right)\left(1+\dfrac{y}{z}\right)\)
=\(\dfrac{y}{y+z}\cdot\left(-\dfrac{z}{y}\right)\cdot\dfrac{y+z}{z}=\dfrac{y}{z}\cdot\left(-\dfrac{z}{y}\right)=-1\)
Vậy A=-1
theo bài ra táo:
\(A=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\\ \Rightarrow A=\dfrac{x-z}{x}.\dfrac{y-x}{y}.\dfrac{z+y}{z}\left(1\right)\)
ta lại có:
\(x-y-z=0\\ \Rightarrow\left\{{}\begin{matrix}x-z=y\left(2\right)\\y-x=-z\left(3\right)\\z+y=x\left(4\right)\end{matrix}\right.\)
thay 2;3;4 vào 1 ta có:
\(A=\dfrac{y}{x}.\dfrac{-z}{y}.\dfrac{x}{z}=-1\)
vậy A = -1
x-y-z=0
\(\Rightarrow x=y+z\)
\(\Rightarrow y=x-z\)
\(\Rightarrow-z=y-z\)
\(B=\left(1-\dfrac{z}{x}\right).\left(1-\dfrac{y}{x}\right).\left(1+\dfrac{y}{z}\right)\)
\(B=\left(\dfrac{x-z}{x}\right).\left(\dfrac{y-x}{y}\right).\left(\dfrac{z+y}{z}\right)\)
\(B=(\dfrac{y}{x}).\left(\dfrac{-z}{y}\right).\left(\dfrac{x}{z}\right)\)
\(B=\dfrac{\left(y.x.-z\right)}{\left(y.x.z\right)}\Rightarrow B=-1\)
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 !!!
ta có x-y-z=0
->x=y+z
y=x-z
z=x-y
B=\(\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1-\dfrac{y}{z}\right)\)
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=\(\dfrac{-\left(xyz\right)}{xyz}\)
B=-1