Question

Bài 4: Cho \(\dfrac{x+y-z}{x}=\dfrac{y+z-x}{y}=\dfrac{z+x-y}{z}\)

Tính A = \(\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)

Answer 1

Từ \(\dfrac{x+y-z}{x}=\dfrac{y+z-x}{y}=\dfrac{z+x-y}{z}\)

=> \(1+\dfrac{y-z}{x}=1+\dfrac{z-x}{y}=1+\dfrac{x-y}{z}\)

=> \(\dfrac{y-z}{x}=\dfrac{z-x}{y}=\dfrac{x-y}{z}\)

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{y-z}{x}=\dfrac{z-x}{y}=\dfrac{x-y}{z}=\dfrac{y-z+z-x+x-y}{x+y+z}=\dfrac{0}{x+y+z}=0\)

Ta có : \(\dfrac{y-z}{x}=0\)

=> y - z = 0 ; Vì x # 0 => y = z

\(\dfrac{z-x}{y}=0\)

=> z - x = 0 . Vì y # 0 => z = x

=> y = z = x

Ta có: A = \(\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)

A = (1 + 1) (1 + 1) ( 1 + 1)

A = 2 . 2 . 2 = 8