Question

cho : \(\dfrac{y+z-2015x}{x}\)=\(\dfrac{z+x-2015y}{y}\)=\(\dfrac{x+y-2015z}{z}\)

tính P= (1+\(\dfrac{x}{y}\))(1+\(\dfrac{y}{z}\))(1+\(\dfrac{z}{x}\))

Answer 1

ta có :\(\dfrac{y+z-2015x}{x}=\dfrac{z+x-2015y}{y}=\dfrac{z+y-2015z}{z}\)

=>\(\left(\dfrac{y+z-2015}{x}+2016\right)=\left(\dfrac{z+x-2015y}{y}+2016\right)=\left(\dfrac{x+y-2015z}{z}+2016\right)\)

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

*Nếu x+y+z\(\ne\)0

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

=>\(P=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)=1.1.1=1

*Nếu x+y+z=0

=>x=y=z

=>\(P=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)=2.2.2=8