Question

[Vũ Ngọc Minh Anh]

y+z-x/x=z+x-y/y=x+y-z/z . Hãy tính giá trị biểu thức B =(1+x/y)(1+y/z)(1+z/x)

Answer 1

[soyeon_Tiểubàng giải]

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

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

+) x + y + z \(\ne0\)

Áp dụng t/c của dãy tỉ số = nhau ta có:

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

\(\Rightarrow\left\{\begin{matrix}y+z-x=x\\z+x-y=y\\x+y-z=z\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}y+z=2x\\z+x=2y\\x+y=2z\end{matrix}\right.\)

\(B=\frac{x+y}{y}.\frac{y+z}{z}.\frac{z+x}{x}=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}=8\)

Answer 2

[Vũ Ngọc Minh Anh]

thank bn