C1:
\(\text{Ta có: }\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{15};x+y+z=6\left(1\right)\)
\(\text{Áp dụng tính dãy tỉ số bằng nhau: }\)
\(\Rightarrow\frac{x+11}{13}=\frac{y+12}{14}=\frac{z+13}{15}=\frac{\left(x+11\right)+\left(y+12\right)+\left(z+13\right)}{13+14+15}=\frac{x+11+y+12+z+13}{42}=\frac{\left(x+y+z\right)+\left(11+12+13\right)}{42}\left(2\right)\)
\(\text{Thay (1) và (2), ta được: }\)
\(\frac{6+36}{42}=\frac{42}{42}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x+11}{13}=1\\\frac{y+12}{14}=1\\\frac{z+13}{15}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+11=13\\y+12=14\\z+13=15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=2\\z=2\end{matrix}\right.\)
\(\text{Vậy }x=2;y=2;z=2\)
C2:
\(x+y+z=6\left(1\right)\)
\(\text{Ta có: }\frac{x+11}{13}=\frac{y+12}{14}\Rightarrow x+11=\frac{y+12}{14}.13\Rightarrow x=\frac{13.\left(y+12\right)}{14}-11\)
\(\frac{y+12}{14}=\frac{z+13}{15}\Rightarrow z+13=\frac{y+12}{14}.15\Rightarrow z=\frac{15.\left(y+12\right)}{14}-13\)
\(\text{Thay }x=\frac{13.\left(y+12\right)}{14}-11;z=\frac{15.\left(y+12\right)}{14}-13\text{ vào }\left(1\right)\)
\(\Rightarrow\frac{13.\left(y+12\right)}{14}-11+y+\frac{15.\left(y+12\right)}{14}-13=6\)
\(\Leftrightarrow\frac{13.\left(y+12\right)}{14}+y+\frac{15.\left(y+12\right)}{14}-\left(11+13\right)=6\)
\(\Leftrightarrow\frac{13.\left(y+12\right)}{14}+\frac{14y}{14}+\frac{15.\left(y+12\right)}{14}-24=6\)
\(\Leftrightarrow\frac{13.\left(y+12\right)+14y+15.\left(y+12\right)}{14}=6+24\)
\(\Leftrightarrow\frac{\left(y+12\right).\left(13+15\right)+14y}{14}=30\)
\(\Leftrightarrow\left(y+12\right).28+14y=30.14\)
\(\Leftrightarrow14.\left[\left(y+12\right).2+y\right]=420\)
\(\Leftrightarrow2y+12.2+y=420:14\)
\(\Leftrightarrow3y+24=30\)
\(\Leftrightarrow3y=30-24\)
\(\Leftrightarrow3y=6\)
\(\Leftrightarrow y=6:3\)
\(\Leftrightarrow y=2\)
\(\text{Khi đó: }x=\frac{13.\left(2+12\right)}{14}-11\left(\text{Do y=2}\right)\)
\(\Leftrightarrow x=\frac{13.14}{14}-11\)
\(\Leftrightarrow x=13-11\)
\(\Leftrightarrow x=2\)
\(\text{Khi đó: }z=\frac{15.\left(2+12\right)}{14}-13\left(\text{Do y=2}\right)\)
\(\Leftrightarrow z=\frac{15.14}{14}-13\)
\(\Leftrightarrow z=15-13\)
\(\Leftrightarrow z=2\)
\(\text{Vậy }x=2;y=2;z=2\)
