hiệu x-y là, nếu\(\frac{x+3}{y+5}=\frac{x+5}{y+7}\)
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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.
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Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x+3}{y+5}=\frac{x+5}{y+7}=\frac{x+3-x-5}{y+5-y-7}=\frac{-2}{-2}=1\)
Suy ra: \(\frac{x+3}{y+5}=1\Rightarrow x+3=y+5\Rightarrow x-y=2\)
Vậy x-y=2
\(\frac{x+3}{y+5}=\frac{x+5}{y+7}\)
=> (x + 3)(y + 7) = (x + 5)(y + 5)
<=> x(y + 7) + 3(y + 7) = x(y+5) + 5(y+5)
<=> xy + 7x + 3y + 21 = xy + 5x + 5y + 25
Cùng bớt xy + 5x + 3y + 21 được :
2x = 2y + 4
=> 2.(x - y) = 4
=> x - y = 2
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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
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Ta có:
\(\frac{x+3}{y+5}=\frac{x+5}{y+7}\)
\(\Rightarrow\left(x+3\right)\left(y+7\right)=\left(x+5\right)\left(y+5\right)\)
\(\Rightarrow xy+7x+3y+21=xy+5x+5y+25\)
\(\Rightarrow\left(7x-5x\right)+\left(3y-5y\right)=25-21\)
\(\Rightarrow2x-2y=4\)
\(\Rightarrow2\left(x-y\right)=4\)
\(\Rightarrow x-y=2\)
Thử lại: Do \(x-y=2\Rightarrow x=y+2\) nên ta có:
\(\frac{\left(y+2\right)+3}{y+5}=\frac{\left(y+2\right)+5}{y+7}\)
\(\Rightarrow\frac{y+5}{y+5}=\frac{y+7}{y+7}\)
\(\Rightarrow1=1\) ( thoả mãn )
Vậy hiệu giữa x và y là 2
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\(\Rightarrow\left(x+3\right)\left(y+7\right)=\left(x+5\right)\left(y+5\right)\)
\(\Rightarrow xy+7x+3y+21=xy+5x+5y+25\)
\(\Rightarrow2x-2y=4\)
\(\Rightarrow2\left(x-y\right)=4\)
\(\Rightarrow x-y=2\)
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\(\frac{x+3}{y+5}=\frac{x+5-x-3}{y+7-y-5}\frac{2}{2}=1\)
x + 3 = y + 5 => x - y = 5 - 3 = 2
vậy x - y = 2
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\(\frac{x+3}{y+5}=\frac{x+5}{y+7}=\frac{x+5-x-3}{y+7-y-5}=\frac{2}{2}=1\)
=> x+3 = y+5
=> x -y = 5 -3 =2
Vậy x -y =2
\(=>\frac{x+5-\left(x+3\right)}{y+7-\left(y+5\right)}=\frac{x+5-x-3}{y+7-y-5}=\frac{2}{2}=1\)
từ x+3/y+5=1
=>x+3=y+5
=>x-y=5-3=2
vậy x-y=2
tikc nhé