Question

C= (1+\(\dfrac{x}{y}\)) . (1+\(\dfrac{x}{z}\)) (1+\(\dfrac{z}{x}\)) biết x+y+z =0

Answer 1

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

Và \(C=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)

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

\(=\dfrac{-z}{y}\cdot\dfrac{-x}{z}\cdot\dfrac{-y}{z}=-\left(\dfrac{xyz}{xyz}\right)=-1\)

Answer 2

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

\(C=\left(\dfrac{y}{y}+\dfrac{x}{y}\right)\left(\dfrac{z}{z}+\dfrac{x}{z}\right)\left(\dfrac{x}{x}+\dfrac{z}{x}\right)\)

\(C=\left(\dfrac{y+x}{y}\right)\left(\dfrac{z+x}{z}\right)\left(\dfrac{x+z}{x}\right)\)

\(C=\dfrac{\left(y+x\right)\left(z+x\right)\left(x+z\right)}{xyz}\)

Rồi ko bt j nx -.-