Hiệu của x - y biết :
\(\frac{x+3}{y+5}=\frac{x+5}{y+7}\)
Hiệu x - y biết:
\(\frac{x+3}{y+5}=\frac{x+5}{y+7}\)
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.
hiệu x - y biết \(\frac{x+3}{y+5}=\frac{x+5}{y+7}\)
\(\Rightarrow\left(x+3\right)\left(y+7\right)=\left(x+5\right)\left(y+5\right)\)
\(\Rightarrow xy+7x+3y+21=xy+5x+5y+25\)
\(\Rightarrow2x-2y=4\)
\(\Rightarrow2\left(x-y\right)=4\)
\(\Rightarrow x-y=2\)
Hiệu x-y biết
\(\frac{x+3}{y+5}=\frac{x+5}{y+7}\)
\(\frac{x+3}{y+5}=\frac{x+5-x-3}{y+7-y-5}\frac{2}{2}=1\)
x + 3 = y + 5 => x - y = 5 - 3 = 2
vậy x - y = 2
tìm hiệu x-y biết: \(\frac{x+3}{y+5}=\frac{x+5}{y+7}\)
\(\frac{x+3}{y+5}=\frac{x+5}{y+7}=\frac{x+5-x-3}{y+7-y-5}=\frac{2}{2}=1\)
=> x+3 = y+5
=> x -y = 5 -3 =2
Vậy x -y =2
\(\Rightarrow\left(x+3\right)\left(y+7\right)=\left(x+5\right)\left(y+5\right)\)
\(\Rightarrow xy+7x+3y+21=xy+5x+5y+25\)
\(\Rightarrow2x-2y=4\Rightarrow x-y=2\)
\(\frac{x+3}{y+5}=\frac{x+5}{y+7}\)
=>(x+3).(y+7)=(x+5).(y+5)
=>x.y+7x+3y+21=xy+5x+5y+25
=>2x-2y=4
=>2(x-y)=4
=>x-y=4:2
=>x-y=2
hiệu x -y biết :
\(\frac{x+3}{y+5}=\frac{x+5}{y+7}\)
1) Hiệu x-y biết :\(\frac{X+3}{Y+5}=\frac{X+5}{Y+7}\)
2) Giá trị của X biết : \(\left(\frac{2}{7}\right)^{6x-7}=1\)
=>x+3 =y+5
=> x -y =5-3
x-y =2 ( do mạng lag = thóat ra)
b) => 6x -7 =0 => x = 7/6
hiệu x-y biết \(\frac{x+3}{y+5}=\frac{x+5}{y+7}\)là
Hiệu x - y biết:
\(\frac{x+3}{y+5}=\frac{x+5}{y+7}\)
=>(x+3)(y+7)=(y+5)(x+5)
x(y+7)+3(y+7)=y(x+5)+5(x+5)
xy+7x+3y+21=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
Hiệu của x-y biết:
\(\frac{x+3}{y+5}\)=\(\frac{x+5}{y+7}\) trả lời nhanh giúp mình nha
=> \(\left(x+3\right)\left(y+7\right)=\left(y+5\right)\left(x+5\right)\)
=> \(xy+7x+3y+3\cdot7=xy+5y+5x+5\cdot5\)
=> \(xy+7x+3y+21=xy+5y+5x+25\)
=> \(xy+7x+3y+21-xy-5y-5x-25=0\)
=> \(xy+7x+3y-xy-5y-5x=-21+25\)
=> \(2x-2y=4\)
=> \(2\left(x-y\right)=4\)
=> \(x-y=4:2\)
=> \(x-y=2\)
tick cho mình nha