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có: \(\frac{x-1}{2005}=\frac{3-y}{2006}=\frac{x-1+3-y}{2005+2006}=\frac{\left(x-y\right)+\left(3-1\right)}{4011}=\frac{4009+2}{4011}=1.\)
=> (x-1)/2005 = 1 => x-1 = 2005 => x = 2006
(3-y)/2006 = 1 => 3-y = 2006 => y = -2003
KL:...\
Áp dụng dãy tỉ số bằng nhau ta có
\(\frac{x-1}{2005}=\frac{3-y}{2006}=\frac{x-1+3-y}{4011}=\frac{4009+2}{4011}=1\)=1
=>x-1=2005<=>x=2006
3-y=2006<=>y=-2003
Chúc học tốt!!!
áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-1}{2005}=\frac{3-y}{2006}\)\(\Leftrightarrow\frac{x-1}{2005}=\frac{3-y}{2006}=\frac{\left(x-1\right)-\left(3-y\right)}{2005-2006}\)\(=\frac{\left(x-y\right)-\left(1+3\right)}{-1}=\frac{4009-4}{-1}=-4005\)
\(\frac{x-1}{2005}=-4005\Rightarrow x=...\left(-4005.2005+1\right)\)
\(\frac{3-y}{2006}=-4005\Rightarrow y=...\left(3-\left(-4005.2006\right)\right)\)
Đặt \(\frac{x-1}{2005}=\frac{3-y}{2006}=k\left(k\ne0\right)\)
\(\Rightarrow x=2005k+1\); \(y=3-2006k\)
Thay x và y vào điều kiện \(x-y=4009\)ta được
\(\left(2005k+1\right)-\left(3-2006k\right)=4009\)
\(\Rightarrow2005k+1-3+2006k=4009\)
\(\Rightarrow2005k+2006k=4009+3-1\)
\(\Rightarrow4011k=4011\)\(\Rightarrow k=1\)
\(\Rightarrow x=2005.1+1=2006\); \(y=3-2006.1=-2003\)
Vậy \(x=2006\); \(y=-2003\)
x-1/2005 = 3-y/2006 =>x-1+3-y/2005+2006=x-y+2/4011=4009+2/4011=1