\(\frac{x-1}{2005}=\frac{3-y}{2006}vax-y=4009\)
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\(\frac{x-1}{2005}=\frac{3-y}{2006}=\frac{x-1+3-y}{2005+2006}=\frac{\left(x-y\right)+\left(-1+3\right)}{2005+2006}=\frac{4009+2}{4011}=\frac{4011}{4011}=1\)
từ x-1/2005=1=>x-1=2005=>x=2006
3-y/2006=1=>3-y=2006=>y=-2003
tick nhé
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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!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x-1}{2005}=\frac{3-y}{2006}=\frac{\left(x-1\right)+\left(3-y\right)}{2005+2006}=\frac{x-y+2}{4011}=\frac{4009+2}{4011}=1\)
\(\Leftrightarrow x-1=2005\Leftrightarrow x=2006\)
\(\Leftrightarrow3-y=2006\Leftrightarrow y=3-2006=-2003\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x-1}{2005}=\frac{3-y}{2006}=\frac{x-1+3-y}{2005+2006}=\frac{x-y-1+3}{4011}=\frac{4009-1+3}{4001}=\frac{4011}{4011}=1\)
=>x-1/2005=1=>x-1=2005=>x=2006
=>3-y/2006=1=>3-y=2006=>y=-2003
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ: \(\frac{x-1}{2015}\)=\(\frac{3-y}{2016}\)=\(\frac{x-1+3-y}{2015+2016}\)=\(\frac{x-y+2}{4011}\)=\(\frac{4009+2}{4011}\)=\(\frac{4011}{4011}\)=1
Do:\(\frac{3-y}{2016}\) = 1 => x=2015
x =2015+ 1
x=2016
\(\frac{3-y}{2016}\)= 1 => y =2016
y=3- 2016
y= -2013
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Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-1}{2005}=\frac{3-y}{2006}=\frac{x-1+3-y}{2005+2006}=\frac{x-y-1+3}{4011}=\frac{4009+2}{4011}=1\)
+) \(\frac{x-1}{2005}=1\Rightarrow x=2006\)
+) \(\frac{3-y}{2006}=1\Rightarrow y=-2003\)
Vậy x = 2006; y = -2003
![](https://rs.olm.vn/images/avt/0.png?1311)
(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?
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