\(a)\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+x+x+z+2+x+y-3}{x+y+z}\)
\(=\dfrac{\left(x+y+z\right)+\left(x+y+z\right)+\left(1+2-3\right)}{x+y+z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)
Lại có: \(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
\(\Rightarrow2=\dfrac{1}{x+y+z}\Rightarrow2\left(x+y+z\right)=1\Rightarrow x+y+z=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z+1}{x}=2\\\dfrac{x+z+2}{y}=2\\\dfrac{x+y-3}{z}=2\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y+z+x+1=3x\\x+y+z+2=3y\\x+y+z-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}+1=3x\\\dfrac{1}{2}+2=3y\\\dfrac{1}{2}-3=3z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1+\dfrac{1}{2}}{3}\\y=\dfrac{\dfrac{1}{2}+2}{3}\\z=\dfrac{\dfrac{1}{2}-3}{3}\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{-5}{6}\end{matrix}\right.\)
Chúc bạn học tốt!
