Question

Cho x, y, z ≠0 và x+y-z=0. Tính A= (1-\(\frac{z}{x}\)).(1-\(\frac{y}{z}\)).(1+\(\frac{x}{y}\))

Answer 1

x + y - z = 0

⇒ x = z - y ; y = z - x ; z = x + y

\(A=\left(1-\frac{z}{x}\right)\left(1-\frac{y}{z}\right)\left(1+\frac{x}{y}\right)\)

\(=\frac{x-z}{x}.\frac{z-y}{z}.\frac{y+x}{y}=\frac{-y}{x}.\frac{x}{z}.\frac{z}{y}=-1\)

Answer 2

\(A=\left(1-\frac{z}{x}\right)\left(1-\frac{y}{z}\right)\left(1+\frac{x}{y}\right)=\frac{x-z}{x}.\frac{z-y}{z}.\frac{x+y}{y}\)

\(x+y-z=0\Leftrightarrow\left\{{}\begin{matrix}x+y=z\\x-z=-y\\z-y=x\end{matrix}\right.\)

thay và A ta được

\(A=-\frac{y}{x}.\frac{z}{y}.\frac{x}{z}=\frac{x.\left(-y\right).z}{x.y.z}=-1\)

vậy A = - 1