\(y=\frac{2x-3}{x-2}=\frac{2x-4+1}{x-2}=\frac{2\left(x-2\right)+1}{x-2}=2+\frac{1}{x-2}\)

Để y nguyên thì 1/x+2 nguyên

hay x + 2 ∈ Ư(1) = { 1 ; -1 }

=> x ∈ { -1 ; -3 }

Vậy ...

Bỏ bài xuống đây nhaa

1 + 1 = 2

xin hết

\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

=> \(\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)

=> \(\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)

=> \(\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

=> \(\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)

=> x - 2010 = 0 (Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\))

=> x = 2010 

Vậy x = 2010

<=>(x-1/2009) + (x-2/2008 -2)=x-3/2007 + (x-4/2006 -2)

<=>(x-1/2009 -1) + (x-2/2008 -1)=(x-3/2007 -1)+ (x-4/2006 -1)

<=>(x-1/2009 -1) + (x-2/2008 -1)=(x-3/2007 -1)+ (x-4/2006 -1)

<=>(x-2010/2009)+(x-2010/2008)-(x-2010/2007)-(x-2010/2006)=0

<=>(x-2010)(1/2009+1/2008-1/2007-1/2006)=0

<=>x-2010=0

<=>x=2010

a, Ta có : \(\left|x-1\right|=2x+1\)

\(\orbr{\begin{cases}x-1=2x+1\\x-1=-2x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}-x=2\\3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=0\end{cases}}}\)

* Trường hợp 1 : \(y=3x-1\Leftrightarrow y=6-1=5\)

* Trường hợp 2 : \(y=2x+1\Leftrightarrow y=0+1=1\)

b, Theo bài ra ta có : \(f\left(3x-1\right)=-1\)hay 

\(3.\left(-1\right)-1=-3-1=-4\)