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

12x - 4y = 3x + 3y

12x - 4y - 3x - 3y = 0

9x - 7y = 0

9x = 7y

x/7 =y/9

=> x/y = 7/9

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

12x-4y=3x+3y

12x-3x=4y+3y

9x=7y

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

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

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

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

=>9x=7y

=>x/y=7/9

Áp dụng tính chất của tỉ lệ thức ta có:

\(\frac{3x-y}{x+y}=\frac{3}{4}\Rightarrow4\left(3x-y\right)=3\left(x+y\right)\)

                    \(\Rightarrow12x-4y=3x+3y\)   

                    \(\Rightarrow12x+3x=3y+4y\)   

                    \(\Rightarrow15x=7y\)     

                   \(\Rightarrow\frac{x}{7}=\frac{y}{15}\Rightarrow\frac{x}{y}=\frac{7}{15}\)       

Vậy tỉ số \(\frac{x}{y}=\frac{7}{15}\)

\(\Rightarrow xy.yz.xz=\left(xyz\right)^2=\frac{1}{3}.\frac{-2}{5}.\frac{-3}{10}=\frac{1}{25}\Rightarrow xyz=\frac{1}{5};\frac{-1}{5}\)

xét xyz=-1/5=>x=1/2;y=2/3;z=-3/5

xét xyz=1/5=>x=-1/2;y=-2/3;z=3/5

Vậy (x;y;z)=(1/2;2/3;-3/5);(-1/2;-2/3;3/5)

Ta có:\(\frac{2x-y}{x+y}=\frac{2}{3}\)

\(\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)

\(\Rightarrow6x-3y=2x+2y\)

\(\Rightarrow4x=5y\)

\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)

\(\frac{3x+2y}{4x-y}=\frac{3}{4}\)

<=>\(4\left(3x+2y\right)=3\left(4x-y\right)\)

<=>12x+8y=12x-3y

<=>12x-12x+8y+3y=0

<=>11y=0

=>y=0 và x=1

1. \(\frac{x}{y}=\frac{7}{17}\)

3. Có 6 cặp

4. 0 có cặp nào hết

Câu 2 mình không biết nha. Thông cảm

\(\frac{2x-y}{x+y}=\frac{2}{3}\Rightarrow\frac{2x-y}{2}=\frac{x+y}{3}=\frac{\left(2x-y\right)-\left(x+y\right)}{2-3}=2y-x\)

\(\Rightarrow2x-y=4y-2x\Rightarrow4x=5y\Rightarrow\frac{x}{y}=\frac{5}{4}\)

Áp dụng công thức lớp 7 ; \(\frac{a}{b}\)= \(\frac{c}{d}\) thì \(\frac{a}{c}\)= \(\frac{b}{d}\)

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

=> .....................................x/y = 5/4

Ta có: \(\frac{2x-y}{x+y}\)=\(\frac{2}{3}\)

=> (2x - y).3 = (x+y) .2

6x - 3y = 2x + 2y

6x - 2x = 3y + 2y

4x = 5y

=> \(\frac{x}{5}\)=\(\frac{y}{4}\)

Vậy tỉ số \(\frac{x}{y}\)=\(\frac{5}{4}\)

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

\(\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)

\(\Rightarrow6x-3y=2x+2y\)

\(\Rightarrow6x-2x=2y+3y\)

\(\Rightarrow4x=5y\)

\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)

Vậy \(\frac{x}{y}=\frac{5}{4}\)

Theo bài ra ta có: \(\frac{2x-y}{x+y}=\frac{2}{3}\)

=> 3(2x-y)=2(x+y)

=> 6x-3y=2x+2y

=> 6x-2x=2y+3y

=> 4x=5y

=> \(\frac{x}{y}=\frac{5}{4}\)

Vậy \(\frac{x}{y}=\frac{5}{4}\)