Question

cho

\(\dfrac{9x}{4}=\dfrac{4y}{3}=\dfrac{12z}{5}\) tính giá trị của biểu thức \(\dfrac{y+z-x}{x-y+z}\)

Answer 1

Ta có \(\dfrac{9x}{4}=\dfrac{4y}{3}=\dfrac{12z}{5}\)

\(\Rightarrow\dfrac{x}{\dfrac{4}{9}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{\dfrac{5}{12}}=\dfrac{y+z-x}{\dfrac{3}{4}+\dfrac{5}{12}-\dfrac{4}{9}}=\dfrac{x-y+z}{\dfrac{4}{9}-\dfrac{3}{4}+\dfrac{5}{12}}\)

\(\Rightarrow\dfrac{y+z-x}{x-y+z}=\dfrac{\dfrac{3}{4}+\dfrac{5}{12}-\dfrac{4}{9}}{\dfrac{4}{9}-\dfrac{3}{4}+\dfrac{5}{12}}=\dfrac{13}{2}\)