Question

\(\frac{x}{3}=\frac{y}{4}và\frac{y}{6}=\frac{z}{5}va3x-2y+5z=86\)

\(\frac{x}{5}=\frac{y}{7};xy=140\)

\(\frac{x-1}{9}+\frac{x-2}{8}=\frac{x-3}{7}+\frac{x-4}{6}\)

\(\frac{31-2x}{x+23}=\frac{9}{4}\)

\(4x=5y;xy-80=0\)

\(\frac{x+3}{8}-\frac{2}{x-3}\)

\(\frac{x^2}{6}=\frac{14}{25}\)

Answer 1

Ta có : \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\left(1\right)\)

           \(\frac{y}{6}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{10}\left(2\right)\)

Từ (1) và (2) => \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}\)

Ta có : \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}=\frac{3x}{27}=\frac{2y}{24}=\frac{5z}{50}=\frac{3x-2y+5z}{27-24+50}=\frac{86}{53}\) (đề sai)

Answer 2

b) Đặt : k = \(\frac{x}{5}=\frac{y}{7}\)

=> k² \(=\frac{x}{5}.\frac{y}{7}=\frac{xy}{35}=\frac{140}{35}=4\)

=> k = -2;2

+ k = 2 thì \(\frac{x}{5}=2\Rightarrow x=10\)

                 \(\frac{z}{7}=2\Rightarrow z=14\)

+ k = -2 thì \(\frac{x}{5}=2\Rightarrow x=-10\)

                 \(\frac{z}{7}=2\Rightarrow z=-14\)

Vậy................................