a, \(\frac{x}{19}=\frac{y}{5}=\frac{z}{95}\); 5x-y-z=-10

biến đổi: 
\(\frac{x}{19}=\frac{5x}{95}\)

=> \(\frac{x}{19}=\frac{y}{5}=\frac{z}{95}\)

(=) \(\frac{5x}{95}=\frac{y}{5}=\frac{z}{95}\)

= \(\frac{5x-y-z}{95-5-95}\)

= \(\frac{-10}{-5}=2\)

* \(\frac{x}{19}=2\)=> \(x=19.2=38\)

* \(\frac{y}{5}=2\)=> \(y=2.5=10\)

* \(\frac{z}{95}=2\)=> \(z=95.2=190\)

nè Khoa ơi câu b có đề ko zợ?

\(F)\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\) và \(2x-y-z=49\)

Ta có: \(\frac{x}{2}=\frac{y}{4}\implies \frac{x}{10}=\frac{y}{15}\)

           \(\frac{y}{5}=\frac{z}{4}\implies\frac{y}{15}=\frac{z}{12} \)

Suy ra: \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{2x}{20}=\frac{2x-y-z}{20-15-12}=\frac{49}{-7}=-7\)

\(\implies \frac{x}{10}=-7\implies x=-70\)

         \(\frac{y}{15}=-7\implies y=-105\)

         \(\frac{z}{12}=-7\implies z=-84\)

Vậy \(x=-70;y=-105;z=-84\)

\(G) \frac{x}{2}=\frac{y}{4}\) và \(xy=2\)

Ta có: \(\frac{x}{2}=\frac{y}{4}\implies \frac{xy}{2}=\frac{y^2}{4}\)

\(\implies \frac{2}{2}=\frac{y^2}{4}\)

\(\implies y^2=2.4:2=4\)

\(\implies y=2=-2\)

\(+)y=2\implies x=1\)

\(+)y=-2\implies x=-1\)

Vậy có các cặp (x;y) là: \((1;2);(-1;-2)\)

\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left[\left(-3,2\right)+\frac{2}{5}\right]\)

\(\Rightarrow\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left[-\frac{3}{2}+\frac{2}{5}\right]\)

\(\Rightarrow\left|x-\frac{1}{3}\right|+\frac{4}{5}=-\frac{11}{10}\)

\(\Rightarrow\left|x-\frac{1}{3}\right|=-\frac{11}{10}-\frac{4}{5}\)

\(\Rightarrow\left|x-\frac{1}{3}\right|=-\frac{19}{10}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{19}{10}\\x-\frac{1}{3}=-\frac{19}{10}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{67}{30}\\x=-\frac{47}{30}\end{cases}}\)

Bạn ơi còn b,c nữa 

dễ lắm nhưng bây h mình k có thời gian để giải