Ta có: \(2x+1=3y+3=5z-3.\)
\(\Rightarrow2.\left(x+\frac{1}{2}\right)=3.\left(y+1\right)=5.\left(z-\frac{3}{5}\right)\)
\(\Rightarrow\frac{2.\left(x+\frac{1}{2}\right)}{30}=\frac{3.\left(y+1\right)}{30}=\frac{5.\left(z-\frac{3}{5}\right)}{30}.\)
\(\Rightarrow\frac{x+\frac{1}{2}}{15}=\frac{y+1}{10}=\frac{z-\frac{3}{5}}{6}\) và \(x-y+z=1,1.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x+\frac{1}{2}}{15}=\frac{y+1}{10}=\frac{z-\frac{3}{5}}{6}=\frac{x+\frac{1}{2}-y-1+z-\frac{3}{5}}{15-10+6}=\frac{\left(x-y+z\right)+\left(\frac{1}{2}-1-\frac{3}{5}\right)}{11}=\frac{1,1-\frac{11}{10}}{11}=0.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x+\frac{1}{2}}{15}=0\Rightarrow x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\\\frac{y+1}{10}=0\Rightarrow y+1=0\Rightarrow y=-1\\\frac{z-\frac{3}{5}}{6}=0\Rightarrow z-\frac{3}{5}=0\Rightarrow z=\frac{3}{5}\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(-\frac{1}{2};-1;\frac{3}{5}\right).\)
Chúc bạn học tốt!
