1. Tìm x biết :
a) \(\dfrac{x+4}{2001}+\dfrac{x+3}{2002}=\dfrac{x+2}{2003}+\dfrac{x+1}{2004}\)
b)\(\dfrac{x+2016}{11}+\dfrac{x+2017}{12}=\dfrac{x+2010}{5}\dfrac{x+2015}{10}\)
c)\(\dfrac{x+3}{5}+\dfrac{x+4}{4}+\dfrac{x+5}{3}+\dfrac{x+12}{2}=-1\)
d) (x-7)(x-6)=0
e)\(^{x^{ }2=4x}\)
f) (x-8)(x+2) đạt giá trị âm
g) \(\dfrac{x-2}{x-4}đạtgiatridương\)
\(\dfrac{x+4}{2001}+\dfrac{x+3}{2002}=\dfrac{x+2}{2003}+\dfrac{x+1}{2004}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2001}+1\right)+\left(\dfrac{x+3}{2002}+1\right)=\left(\dfrac{x+2}{2003}+1\right)+\left(\dfrac{x+1}{2004}+1\right)\)
\(\Leftrightarrow\dfrac{x+2005}{2001}+\dfrac{x+2005}{2002}-\dfrac{x+2005}{2003}-\dfrac{x+2005}{2004}=0\)
\(\Leftrightarrow\left(x+2005\right)\cdot\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)=0\)
Mà \(\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)\ne0\)
\(\Rightarrow x+2005=0\Rightarrow x=-2005\)
