Question

[Phạm Thị Hằng]

Cho ba số x ; y ; z khác 0 thỏa mãn điều kiện \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\) 
Khi đó B = \(\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\) có giá trị bằng .......

Answer 1

[Hoàng Phúc]

theo t/c dãy tỉ số=nhau:

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

=>x=y=z

\(1+\frac{x}{y}=\frac{x+y}{y}=\frac{y+y}{y}=\frac{2y}{y}=2\)

\(1+\frac{y}{z}=\frac{y+z}{z}=\frac{z+z}{z}=\frac{2z}{z}=2\)

\(1+\frac{z}{x}=\frac{z+x}{x}=\frac{x+x}{x}=\frac{2x}{x}=2\)

=>B=2.2.2=8

Answer 2

[Võ Đông Anh Tuấn]

\(\frac{3x+3y+3z}{x+y+z}\)=\(\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{2};y=\frac{1}{2};z=-\frac{1}{2}\)

\(\Leftrightarrow B=\left(1+\frac{\frac{1}{2}}{\frac{1}{2}}\right)\left(1+\frac{\frac{1}{2}}{\frac{-1}{2}}\right)\left(1+\frac{\frac{-1}{2}}{\frac{1}{2}}\right)\)=0