Question

Cho \(x,y,z\ne0\) và \(x-y-z=0\).

Tính giá trị biểu thức \(A=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\)

Answer 1

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

Answer 2

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