Question

tìm x;y biết:

\(\frac{x-1}{2005}=\frac{3-y}{2006}\)  và x-y=4009

Answer 1

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:...\

Answer 2

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!!!

Answer 3

áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

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

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

Answer 4

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 \(x=2006\); \(y=-2003\)

Answer 5

bằng -2003

Answer 6

x-1/2005 = 3-y/2006 =>x-1+3-y/2005+2006=x-y+2/4011=4009+2/4011=1