Đáp án:

Giải thích các bước giải:

$\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$

$⇔\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0$

$⇔(x+1)\left (\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14} \right )=0$

$⇔x+1=0$ Vì $\left (\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14} \right ) \neq 0$

$⇔x=-1$

$\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$

$⇔\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1$

$⇔\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$

$⇔\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0$

$⇔(x+2004)\left (\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003} \right )=0$

$⇔x+2004=0$ Vì $\left (\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003} \right ) \neq 0$

$⇔x=-2004$

Xin hay nhất nhé.

$a)$

$\frac{x+1}{10}+$ $\frac{x+1}{11}+$ $\frac{x+1}{12}=$ $\frac{x+1}{13}+$ $\frac{x+1}{14}$ 
$⇔\frac{x+1}{10}+$ $\frac{x+1}{11}+$ $\frac{x+1}{12}-$ $\frac{x+1}{13}-$ $\frac{x+1}{14}=0$ 
$⇔(x+1)^{}.$ $(\frac{1}{10}+$ $\frac{1}{11}+$ $\frac{1}{12}-$$\frac{1}{13}-$$\frac{1}{14})=0$
$⇔x+1^{}=0$ $(vì$ $\frac{1}{10}+$ $\frac{1}{11}+$ $\frac{1}{12}-$$\frac{1}{13}-$$\frac{1}{14}$ $\neq0)$
$⇔x=-1^{}$ 

                  $Vậy$ $x=-1$

$b)$

$\frac{x+4}{2000}+$ $\frac{x+3}{2001}=$ $\frac{x+2}{2002}+$ $\frac{x+1}{2003}$ 
$⇔\frac{x+4}{2000}+1+$ $\frac{x+3}{2001}+1=$ $\frac{x+2}{2002}+1+$ $\frac{x+1}{2003}+1$
$⇔\frac{x+2004}{2000}+$ $\frac{x+2004}{2001}=$ $\frac{x+2004}{2002}+$ $\frac{x+2004}{2003}$
$⇔\frac{x+2004}{2000}+$ $\frac{x+2004}{2001}$ $-\frac{x+2004}{2002}$ $-\frac{x+2004}{2003}=0$
$⇔(x+2004)^{}.$ $(\frac{1}{2000}+$ $\frac{1}{2001}-$ $\frac{1}{2002}-$$\frac{1}{2003})=0$
$⇔x+2004^{}=0$$(vì$$\frac{1}{2000}+$ $\frac{1}{2001}-$ $\frac{1}{2002}-$$\frac{1}{2003}$ $\neq0)$
$⇔x=-2004^{}$ 

              $Vậy$ $x=-2004$