1.Tìm x, biết.
a, \(|x|\) + x = \(\dfrac{1}{3}\)
b, \(|x+2|\) = x
c, \(\dfrac{2}{\left(x-1\right)\left(x-3\right)}\) + \(\dfrac{5}{\left(x-3\right)\left(x-8\right)}\) + \(\dfrac{12}{\left(x-8\right)\left(x-20\right)}\) - \(\dfrac{1}{x-20}\) = \(\dfrac{-3}{4}\)
d, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}\) + \(\dfrac{5}{\left(x+5\right)\left(x+10\right)}\) + \(\dfrac{7}{\left(x+10\right)\left(x+17\right)}\) = \(\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
e, \(\dfrac{x-1}{2009}\) + \(\dfrac{x-2}{2008}\) = \(\dfrac{x-3}{20007}\) + \(\dfrac{x-4}{2006}\)
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
