Question

Tìm các số hữu tỉ x,y,z biết rằng \(\dfrac{4}{x+1}=\dfrac{2}{y-2}=\dfrac{3}{z+2}\) và \(2y^2-\left(z+5\right)^2=-25\)

Answer 1

\(\dfrac{4}{x+1}=\dfrac{2}{y-2}=\dfrac{3}{z+2}\Rightarrow\dfrac{x+1}{4}=\dfrac{y-2}{2}=\dfrac{z+2}{3}\)

\(\Rightarrow\dfrac{x+5}{4}=\dfrac{y}{2}=\dfrac{z+5}{3}\Rightarrow\dfrac{\left(x+5\right)^2}{16}=\dfrac{y^2}{4}=\dfrac{\left(z+5\right)^2}{9}\)

\(\Rightarrow\dfrac{\left(x+5\right)^2}{16}=\dfrac{2y^2}{8}=\dfrac{\left(z+5\right)^2}{9}=\dfrac{2y^2-\left(z+5\right)^2}{8-9}=\dfrac{-25}{-1}=25\)

\(\Rightarrow\left(x+5\right)^2=16.25\Rightarrow x+5=\pm20\)

Nếu \(x=15\Rightarrow y=\dfrac{x+5}{2}=10;z=\dfrac{3\left(x+5\right)}{4}-5=10\)

Nếu \(x=-25\Rightarrow y=\dfrac{x+5}{2}=-10;z=\dfrac{3\left(x+5\right)}{4}-5=-20\)