\(x^4+4x^2-5=0\)
\(\Leftrightarrow x^4-x^2+5x^2-5=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)+5\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+5\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+5\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5=0\left(l\right)\\x=1\\x=-1\end{matrix}\right.\)
\(4\left(x+5\right)-3\left|2x-1\right|=0\)
\(\Leftrightarrow3\left|2x-1\right|=4\left(x+5\right)\)
\(\Leftrightarrow\left|2x-1\right|=\dfrac{4}{3}\left(x+5\right)\left(ĐK:x\ge-5\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{4}{3}\left(x+5\right)\\2x-1=-\dfrac{4}{3}\left(x+5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{4}{3}x+\dfrac{20}{3}\\2x-1=-\dfrac{4}{3}x-\dfrac{20}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{23}{3}\\\dfrac{2}{3}x=-\dfrac{17}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{2}\left(l\right)\\x=-\dfrac{17}{10}\left(n\right)\end{matrix}\right.\)
Vậy: \(x=-\dfrac{17}{10}\)
\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\\ \Leftrightarrow\dfrac{2-x}{2007}+1=\left(\dfrac{1-x}{2008}+1\right)+\left(1-\dfrac{x}{2009}\right)\\ \Leftrightarrow\dfrac{2009-x}{2007}=\dfrac{2009-x}{2008}+\dfrac{2009-x}{2009}\\ \Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2009}\right)=0\\ \Leftrightarrow2009-x=0\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2009}\ne0\right)\\ \Leftrightarrow x=2019\)
Vậy phương trình có nghiệm \(x=2019\)
