\(\frac{3x-y}{x+y}=\frac{3}{4}\Rightarrow\left(3x-y\right)4=3\left(x+y\right)\)

=>12x-4y=3x+3y

=>12x-4y=3x+3y

=>12x-3x=3y+4y

=>9x=7y

=>x/y=7/9

vậy x/y=7/9

Ta có: \(\frac{3x-y}{x+y}\)=\(\frac{3}{4}\)

\(\Leftrightarrow\)4(3x-y)=3(x+y)

\(\Leftrightarrow\)12x-4y=3x+3y

\(\Leftrightarrow\)12x-3x=4x+3y

\(\Leftrightarrow\)9x=7y

\(\Leftrightarrow\)\(\frac{x}{y}\)=\(\frac{7}{9}\)

\(\frac{3x-y}{x+y}=\frac{3}{4}\Leftrightarrow\frac{3x-y}{x+y}+1=\frac{3}{4}+1\Leftrightarrow\frac{4x}{x+y}=\frac{7}{4}.\) Ở vế trái chia cả tử và mẫu cho y , được:

\(\frac{4.\frac{x}{y}}{\frac{x}{y}+1}=\frac{7}{4}\)  Suy ra :  \(16.\frac{x}{y}=7\left(\frac{x}{y}+1\right)\) Vậy \(9.\frac{x}{y}=7\Leftrightarrow\frac{x}{y}=\frac{7}{9}\)

\(\frac{3x-y}{x+y}=\frac{3}{4}\)

\(=>4\left(3x-y\right)=3\left(x+y\right)\)

\(12x-4y=3x+3y\)

\(12x-3x=3y+4y\)

\(9x=7y\)

\(=>\frac{x}{y}=\frac{7}{9}\)

\(\frac{3x-y}{x+y}=\frac{3}{4}\Rightarrow12x-4y=3x+3y\Rightarrow9x=7y\Rightarrow\frac{x}{y}=\frac{7}{9}\)

từ tỉ lệ thức ta có:

4(3x-y)=3(x+y)

12x-4y=3x+3y

9x-4y=3y

9x=7y

x/y=7/9

Ta có \(\frac{3x-y}{x+y}=\frac{3}{4}\)Suy ra 4(3x-y)=3(x+y)

=>12x-4y=3x+3y

=>12x-4y-3x=3y

=>12x-3x=3y+4y

=>9x=7y=>\(\frac{x}{y}=\frac{9}{7}\)

Nhớ k cho mik nha

\(\frac{3x-y}{x+y}=\frac{3}{4}\Leftrightarrow4\left(3x-y\right)=3\left(x+y\right)\)

\(\Leftrightarrow12x-4y=3x+3y\Leftrightarrow12x-3x=3y+4y\Leftrightarrow9x=7y\)

\(\Leftrightarrow\frac{x}{y}=\frac{7}{9}\)

nhầm, sửa lại:

\(\frac{3x-y}{x+y}=\frac{3}{4}\)

=> 4 (3x - y) = 3 (x + y)

=> 12x - 4y = 3x + 3y

=> 12x - 3x = 3y + 4y

=> 9x = 7y

=> x/y = 7/9

                     Vậy giá trị của tỉ số x/y là 7/9

Có \(\frac{3x-y}{x+y}=\frac{1}{2}\)

=> (3x- y).2 = x + y

=> 6x - 2y = x + y 

=> 5x = 3y

=> \(\frac{x}{y}=\frac{3}{5}\)

\(\frac{3x-y}{x+y}=\frac{1}{2}\)

\(\Rightarrow2\left(3x-y\right)=x+y\)

\(\Rightarrow6x-2y=x+y\)

\(\Rightarrow5x=3y\)

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

Vậy :................