Bài `11:`
`a)\frac{3}{(x+2)(x+5)}+\frac{5}{(x+5)(x+10)}+\frac{7}{(x+10)(x+17)}=\frac{x}{(x+2)(x+17)}` $(*)$
ĐKXĐ: $\begin{cases}x+2\ne0\\x+5\ne0\\x+10\ne0\\x+17\ne0\end{cases}$$\Leftrightarrow\begin{cases}x\ne-2\\x\ne-5\\x\ne-10\\x\ne-17\end{cases}$
$⇒(*)$`⇔\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{(x+2)(x+17)}`
`⇔\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{(x+2)(x+17)}`
`⇔\frac{x+17-x-2}{(x+2)(x+17)}=\frac{x}{(x+2)(x+17)}`
`⇔\frac{15}{(x+2)(x+17)}=\frac{x}{(x+2)(x+17)}`
`⇒x=15` $(tmdk)$
Vậy `x=15.`
`b)\frac{2}{(x-1)(x-3)}+\frac{5}{(x-3)(x-8)}+\frac{12}{(x-8)(x-20)}-\frac{1}{x-20}=-3/4` $(*)$
ĐKXĐ: $\begin{cases}x-1\ne0\\x-3\ne0\\x-8\ne0\\x-20\ne0\end{cases}$$\Leftrightarrow\begin{cases}x\ne1\\x\ne3\\x\ne8\\x\ne20\end{cases}$
$⇒(*)$`⇔\frac{2}{(x-1)(x-3)}+\frac{5}{(x-3)(x-8)}+\frac{12}{(x-8)(x-20)}-\frac{1}{x-20}=-3/4`
`⇔\frac{1}{x-3}-\frac{1}{x-1}+\frac{1}{x-8}-\frac{1}{x-3}+\frac{1}{x-20}-\frac{1}{x-8}-\frac{1}{x-20}=-3/4`
`⇔-\frac{1}{x-1}=-3/4`
`⇔(-1).4=(-3)(x-1)`
`⇔-4=(-3)(x-1)`
`⇔-4/-3=x-1`
`⇔4/3=x-1`
`⇔4/3+1=x`
`⇔x=7/3. `$(tmdk)$
Vậy `x=7/3.`
`c)``\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}`
`⇔\frac{x-1}{2009}+\frac{x-2}{2008}-2=\frac{x-3}{2007}+\frac{x-4}{2006}-2`
`⇔\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1`
`⇔\frac{x-1}{2009}-2009/2009+\frac{x-2}{2008}-2008/2008=\frac{x-3}{2007}-2007/2007+\frac{x-4}{2006}-2006/2006`
`⇔\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`
`⇔(x-2010)(1/2009+1/2008-1/2007-1/2006)=0`
`⇔x-2010=0` (vì `1/2009+1/2008-1/2007-1/2006\ne0`)
`⇔x=0+2010`
`⇔x=2010.`
Vậy `x=2010.`
