\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
Áp dụng t/c ................ ta có :
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{45}{9}=5\)
\(\Rightarrow\frac{2x-2}{4}=5\Rightarrow x=11\)
\(\Rightarrow\frac{3y-6}{9}=5\Rightarrow y=17\)
\(\Rightarrow\frac{z-3}{4}=5\Rightarrow z=23\)
Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\) =>\(\hept{\begin{cases}x=2k+1\\y=3k+2\\z=4k+3\end{cases}}\)
Có: \(2x+3y-z=50\)
<=>\(2\left(2k+1\right)+3\left(3k+2\right)-\left(4k+3\right)=50\)
<=>\(4k+2+9k+6-4k-3=50\)
<=>\(9k+5=50\)<=>\(9k=45\)<=>\(k=5\)
=>\(\hept{\begin{cases}x=2.5+1=11\\y=3.5+2=17\\z=4.5+3=23\end{cases}}\)
Vậy ..............
Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\Leftrightarrow\hept{\begin{cases}x=2k+1\\y=3k+2\\z=4k+3\end{cases}}\)
Theo giả thiết ta có: \(2x+3y-z=2\left(2k+1\right)+3\left(3k+2\right)-\left(4k+3\right)=50\)
\(\Leftrightarrow\left(4k+2\right)+\left(9k+6\right)-\left(4k+3\right)=50\)
\(\Leftrightarrow4k+2+9k+6-4k-3=50\)
\(\Leftrightarrow\left(4k+9k-4k\right)+\left(2+6-3\right)=50\)
\(\Leftrightarrow9k+5=50\Leftrightarrow9k=45\Leftrightarrow k=5\)
Thế vào ta có: \(\hept{\begin{cases}x=2k+1\\y=3k+2\\z=4k+3\end{cases}\Leftrightarrow\hept{\begin{cases}x=11\\y=17\\z=23\end{cases}}}\)
