Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x+3}{y+5}=\frac{x+5}{y+7}=\frac{\left(x+3\right)-\left(x+5\right)}{\left(y-5\right)-\left(y+7\right)}=\frac{x+3-x+5}{y+5-y+7}=\frac{3-5}{5-7}=\frac{-2}{-2}=1\)

\(\Rightarrow\frac{x+3}{y+5}=\frac{x+5}{y+7}=1\)

\(\Rightarrow\left(x+3\right).1=\left(y+5\right).1\)

\(\Rightarrow x+3=y+5\)

\(\Rightarrow x-y=5-3\)

\(\Rightarrow x-y=2\)

Vậy x - y = 2.

Đúng không mài

Dễ mà. 

\(\frac{x+3}{y+5}=\frac{x+5}{y+7}\)

<=> (x+3)(y+7) = (y+5)(x+5)

<=> xy + 7x + 3y + 21 = xy + 5y + 5x + 25

<=> xy - xy + 7x - 5x + 3y - 5y = 25 - 21

<=> 2x - 2y = 4

<=> 2(x-y) = 4

<=> x-y = 4/2 = 2

Vậy x-y = 2.

\(\frac{x+3}{y+5}=\frac{x+5}{y+7}=\frac{x+5-x-3}{y+7-y-5}=\frac{2}{2}=1\)

=> x+3 = y+5

=> x -y = 5 -3 = 2

vậy x -y = 2

Từ \(\frac{x+3}{y+5}=\frac{x+5}{y+7}\)

=>(x+3).(y+7)=(y+5).(x+5)

=>x(y+7)+3(y+7)=y(x+5)+5(x+5)

=>xy+7x+3y+21=yx+5y+5x+25

=>7x+3y+21=5y+5x+25

=>5x+2x+3y+21=2y+3y+5x+21+4

=>2x=2y+4

=>2x-2y=4

=>2(x-y)=4=>x-y=2

VẬy x-y=2

Áp dụng t/c dãy tỉ số bằng nhau : \(\frac{x+3}{y+5}=\frac{x+5}{y+7}=\frac{\left(x+5\right)-\left(x+3\right)}{\left(y+7\right)-\left(y+5\right)}=\frac{2}{2}=1\)

\(\Rightarrow x+3=y+5\Rightarrow x-y=5-3=2\)

Áp dụng tính chất dãy tỉ số bằng nhau :\(\frac{x+3}{y+5}\)=\(\frac{x+5}{y+7}\) =\(\frac{\left(x+5\right)-\left(x+3\right)}{\left(x+7\right)-\left(y+5\right)}\)=\(\frac{2}{2}\)=1

\(\Rightarrow\)x+3 = y +5 

\(\Rightarrow\)x -y = 5-3=2

\(\Rightarrow\left(x+3\right)\left(y+7\right)=\left(x+5\right)\left(y+5\right)\)

\(\Rightarrow xy+7x+3y+21=xy+5x+5y+25\)

\(\Rightarrow2x-2y=4\)

\(\Rightarrow2\left(x-y\right)=4\)

\(\Rightarrow x-y=2\)

\(\frac{x+3}{y+5}=\frac{x+5-x-3}{y+7-y-5}\frac{2}{2}=1\)

x + 3 = y + 5 => x - y = 5 - 3 = 2

vậy x - y = 2

câu hỏi tương tự nha bạn

cảm ơn nha 

Sakura

        x-y=2

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\begin{array}{l}\frac{x}{5} = \frac{y}{7} = \frac{z}{9} = \frac{{x - y + z}}{{5 - 7 + 9}} = \frac{{\frac{7}{3}}}{7} = \frac{7}{3}.\frac{1}{7} = \frac{1}{3}\\ \Rightarrow x = 5.\frac{1}{3} = \frac{5}{3};\\y = 7.\frac{1}{3} = \frac{7}{3};\\z = 9.\frac{1}{3} = \frac{9}{3} = 3.\end{array}\)

Vậy \(x = \frac{5}{3};y = \frac{7}{3};z = 3\)

=>(x+3)(y+7)=(y+5)(x+5)

x(y+7)+3(y+7)=y(x+5)+5(x+5)

xy+7x+3y+21=xy+5y+5x+25

(xy-xy)+(7x-5x)=(5y-3y)+(25-21)

2x=2y+4

2x-2y=4

2(x-y)=4

x-y=4/2

x-y=2