a: (2x+5)(1-4x)=0
=>\(\left[{}\begin{matrix}2x+5=0\\1-4x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=-5\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=\dfrac{1}{4}\end{matrix}\right.\)
b:
ĐKXĐ: \(x\notin\left\{4;-1\right\}\)
\(\dfrac{x+4}{x+1}-\dfrac{x+1}{x-4}=\dfrac{15x}{x^2-3x-4}\)
=>\(\dfrac{x+4}{x+1}-\dfrac{x+1}{x-4}=\dfrac{15x}{\left(x-4\right)\left(x+1\right)}\)
=>\(\dfrac{\left(x+4\right)\left(x-4\right)-\left(x+1\right)^2}{\left(x+1\right)\left(x-4\right)}=\dfrac{15x}{\left(x-4\right)\left(x+1\right)}\)
=>\(x^2-16-x^2-2x-1=15x\)
=>-2x-17=15x
=>-17x=17
=>x=-1(loại)
c: \(13-5x>-3x+9\)
=>-5x+3x>9-13
=>-2x>-4
=>2x<4
=>x<2
d: \(\dfrac{x+1}{3}+\dfrac{2x+1}{4}< =\dfrac{5x+3}{6}+\dfrac{12x+7}{12}\)
=>\(\dfrac{4\left(x+1\right)}{12}+\dfrac{3\left(2x+1\right)}{12}< =\dfrac{2\left(5x+3\right)+12x+7}{12}\)
=>4x+4+6x+3<=10x+6+12x+7
=>10x+7<=22x+13
=>-12x<=6
=>\(x>=-\dfrac{6}{12}=-\dfrac{1}{2}\)
