\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}=\frac{5z-25}{30}=\frac{3x-3}{6}=\frac{4y+12}{16}=\frac{\left(5z-3x-4y\right)-\left(25-3+12\right)}{8}=\frac{50-34}{8}=2\Rightarrow\left\{{}\begin{matrix}x=2.2+1=5\\y=4.2-3=5\\z=6.2+5=17\end{matrix}\right.\)
\(\text{Áp dụng tính chất dãy tỉ số bằng nhau ta có:}\)
\( \dfrac{{x - 1}}{2} = \dfrac{{y + 3}}{4} = \dfrac{{z - 5}}{6} = \dfrac{{3x - 3}}{6} = \dfrac{{4y + 12}}{{16}} = \dfrac{{5z - 25}}{{30}}\\ = \dfrac{{5z - 25 - 3x + 3 - 4y + 12}}{{30 - 6 - 16}} = \dfrac{{\left( {5z - 3y - 4y} \right) - \left( {25 - 3 - 12} \right)}}{8} = \dfrac{{50 - 10}}{8} = \dfrac{{40}}{8} = 5\\ *\dfrac{{x - 1}}{2} = 5 \Rightarrow x = 11\\ *\dfrac{{y + 3}}{4} = 5 \Rightarrow y = 17\\ *\dfrac{{z - 5}}{6} = 5 \Rightarrow z = 35 \)Vậy \((x;y;z)=(11;17;35)\)
Ây da :)) Xin lỗi sai rồi :)) Có thêm cách nữa:v
Ta có: \(\dfrac{x-1}{2} = \dfrac{y+3}{4} = \dfrac{z-5}{6} = k\)
\(x = 2k+1 ; y = 4k+3 ; z = 6k+5\)
Thay các giá trị: \(x = 2k+1 ; y = 4k-3 ; z = 6k+5 \)vào biểu thức
\(5z - 3x - 4y = 50\). Ta có,
\(5.(6k+5) - 3.(4k+3) - 4.(4k-3) = 50\)
\(\Leftrightarrow\) \(30k + 25 - 6k - 3 - 16k+ 12 = 50\)
\(\Leftrightarrow\)\(8k + 34 = 50\)
\(\Leftrightarrow\) \(8k = 50-34 = 16\)
\(\Leftrightarrow\) \(k = \dfrac{16}{8} = 2\)
\(\Rightarrow\) \(\dfrac{x-1}{2} = 2 \) \(\Rightarrow\) \(x=5\)
\(\Rightarrow\)\(\dfrac{y-3}{4} = 2\)\(\Rightarrow y=5\)
\(\Rightarrow\)\(\dfrac{z-5}{6}=2\)\(\Rightarrow z=17\)
Vậy ....