\(\frac{x-1}{2005}=\frac{3-y}{2006}=\frac{x-1+3-y}{2005+2006}=\frac{\left(x-y\right)+\left(-1+3\right)}{2005+2006}=\frac{4009+2}{4011}=\frac{4011}{4011}=1\)

từ x-1/2005=1=>x-1=2005=>x=2006

3-y/2006=1=>3-y=2006=>y=-2003

tick nhé

ADTCDTSBN

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:

\(\frac{x-1}{2005}=\frac{3-y}{2006}\)\(\Leftrightarrow\frac{x-1}{2005}=\frac{3-y}{2006}=\frac{\left(x-1\right)-\left(3-y\right)}{2005-2006}\)\(=\frac{\left(x-y\right)-\left(1+3\right)}{-1}=\frac{4009-4}{-1}=-4005\)

\(\frac{x-1}{2005}=-4005\Rightarrow x=...\left(-4005.2005+1\right)\)

\(\frac{3-y}{2006}=-4005\Rightarrow y=...\left(3-\left(-4005.2006\right)\right)\)

\(\frac{x-1}{2005}=\frac{3-y}{2006}=\frac{x-1+3-y}{2005+2006}=\frac{x-y-1+3}{4011}=\frac{4009-1+3}{4001}=\frac{4011}{4011}=1\)

=>x-1/2005=1=>x-1=2005=>x=2006

=>3-y/2006=1=>3-y=2006=>y=-2003

(x-1)/2005=(3-y)/2006

nên x/2006=y/-2003

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

 x/2006=y/-2003=(x+y)/(2006+2003)=4009/4009=1

nên x=2006

      y=-2003

z ở đâu ra đấy?

mình cần gấp

Áp dụng TC DTSBN ta có :

\(\frac{x-1}{2005}=\frac{3-y}{2006}=\frac{x-1+3-y}{2005+2006}=\frac{x-y+2}{4011}=\frac{4009+2}{4011}=1\)

\(\Rightarrow\frac{x-1}{2005}=1\Rightarrow x-1=2005\Rightarrow x=2006\)

\(\Rightarrow\frac{3-y}{2006}=1\Rightarrow3-y=2006\Rightarrow y=-2003\)

Vậy x = 2006; y = - 2003