a

Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)

\(\Rightarrow x=2k+1;y=3k+2;z=4k+3\)

Thay vào,ta được:

\(2\left(2k+1\right)+3\left(3k+2\right)-\left(4k+3\right)=50\)

\(\Leftrightarrow4k+2+9k+6-4k-3=50\)

\(\Leftrightarrow9k+5=50\)

\(\Leftrightarrow9k=45\)

\(\Leftrightarrow k=5\)

\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}=\frac{5x-5}{10}=\frac{3y+9}{12}=\frac{4z-20}{24}\)

\(=\frac{5x-5-3y-9-4z+20}{10-12-24}=\frac{\left(5x-3y-4z\right)+\left(20-5-9\right)}{26}=\frac{46+6}{26}=2\)

\(\Rightarrow x=2\cdot2+1=5\)

\(y=4\cdot2-3=5\)

\(z=2\cdot6+5=17\)

Câu c tương tự như câu 1

\(c,\frac{2x}{5}=\frac{3y}{10}=\frac{z}{12}\)và x + y + z = 107

Ta có : \(\frac{2x}{5}=\frac{3y}{10}=\frac{z}{12}\Leftrightarrow\frac{x}{\frac{5}{2}}=\frac{y}{\frac{10}{3}}=\frac{z}{12}=\frac{x+y+z}{\frac{5}{2}+\frac{10}{3}+12}=\frac{107}{\frac{107}{6}}=107\cdot\frac{6}{107}=6\)

Vậy : \(\hept{\begin{cases}\frac{2x}{5}=6\\\frac{3y}{10}=6\\\frac{z}{12}=6\end{cases}}\Leftrightarrow\hept{\begin{cases}x=15\\x=20\\z=72\end{cases}}\)

a) ta co: x\5=y\3=z\4 va x+2y-z=-121

      Dat: x\5=y\3=z\4=k.suy ra: x=5k;y=3k;z=4k

                                              =5k+2.(3k)-4k

                                              =5k+6k-4k

                                              =7k=-121

                                              =-121:7k=-121\7

suy ra:x\5=-121\7suy ra: -121\7.5=-605\7

          y\3=-121\7 suy ra:-121\7.3=-363\7

          z\4=-121\7 suy ra:-121\7.3=-484\7

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{3}=\frac{y}{6}=\frac{z}{5}=\frac{2xy-3y^2-4yz}{2.3.6-3.6^2-4.6.5}=\frac{24}{-192}=\left(-\frac{1}{8}\right)\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{3}=\left(-\frac{1}{8}\right)\Rightarrow x=\left(-\frac{3}{8}\right)\\\frac{y}{6}=\left(-\frac{1}{8}\right)\Rightarrow y=\left(-\frac{3}{4}\right)\\\frac{z}{5}=\left(-\frac{1}{8}\right)\Rightarrow z=\left(-\frac{5}{8}\right)\end{cases}}\)

Vậy .... 

\(b,\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}=\)

\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=\frac{2x+3y-5z}{2.2+3.3-5.5}=\frac{\left(-12\right)}{\left(-12\right)}=1\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=1\Rightarrow x=2\\\frac{y}{3}=1\Rightarrow y=3\\\frac{z}{5}=1\Rightarrow z=5\end{cases}}\)

Vậy ...