Question

 Cho \(\frac{y+z}{x}=\frac{z+x}{y}=\frac{x+y}{z}\). Tính \(B=\left(1+\frac{x}{y+z}\right)\left(1+\frac{y}{x+z}\right)\left(1+\frac{z}{x+y}\right)\)

 

Answer 1

áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{y+z}{x}\)=\(\frac{z+x}{y}\)=\(\frac{x+y}{z}\)=\(\frac{2\left(x+y+z\right)}{x+y+z}\)= 2

từ \(\frac{y+z}{x}\)=2 suy ra y+z=2x

từ \(\frac{z+x}{y}\)=2 suy ra z+x=2y

từ \(\frac{x+y}{z}\)=2 suy ra x+y=2z

thay vào ta có:

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

 = (1+1/2)(1+1/2)(1+1/2)

 =3/2.3=9/2