Question

Cho 3 số \(x;y;z\) thỏa mãn \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)và\(2x+3y-z=50\)khi đó \(x+y+z=\)

Answer 1

Ta có: \(\frac{x-1}{2}=\frac{2x-2}{4};\frac{y-2}{3}=\frac{3y-6}{9};\frac{z-3}{4}=\frac{-z+3}{-4}\)

Vì \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)nên \(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{-z+3}{-4}=\frac{2x-2+3y-6-z+3}{4+9+\left(-4\right)}=\frac{50-5}{9}=\frac{45}{9}=5\)

\(\Rightarrow\frac{x-1}{2}=5\Rightarrow x=11\)

\(\Rightarrow\frac{y-2}{3}=5\Rightarrow y=17\)

\(\Rightarrow\frac{z-3}{4}=5\Rightarrow z=23\)

Vậy, \(x+y+z=11+17+23=51\)