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

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

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

\(\Rightarrow2x-2y=4\)

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

\(\Rightarrow x-y=2\)

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

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

vậy x - y = 2

câu hỏi tương tự nha bạn

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

=> x+3 = y+5

=> x -y = 5 -3 =2

Vậy x -y =2

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

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

\(\Rightarrow2x-2y=4\Rightarrow x-y=2\)

        x-y=2

=>x+3 =y+5 

=> x -y =5-3 

x-y =2  ( do mạng lag = thóat ra)

b)  => 6x -7  =0  => x = 7/6

=>(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

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

=> \(xy+7x+3y+3\cdot7=xy+5y+5x+5\cdot5\)

=> \(xy+7x+3y+21=xy+5y+5x+25\)

=> \(xy+7x+3y+21-xy-5y-5x-25=0\)

=> \(xy+7x+3y-xy-5y-5x=-21+25\)

=> \(2x-2y=4\)

=> \(2\left(x-y\right)=4\)

=> \(x-y=4:2\)

=> \(x-y=2\)

tick cho mình nha