a: \(\Leftrightarrow\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}=\dfrac{x-5}{2\left(x-5\right)\left(x+5\right)}\)
=>2(x+5)^2-(x-5)^2=x(x-5)
=>2x^2+20x+50-x^2+10x-25=x^2-5x
=>30x+25=-5x
=>35x=-25
=>x=-5/7
b: =>\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}-\dfrac{3}{\left(x-2\right)\left(x+1\right)}=\dfrac{-1}{x\left(x-4\right)}\)
=>\(\dfrac{x-2-3x-6}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{-1}{x\left(x-4\right)}\)
\(\Leftrightarrow\dfrac{-2x-8}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{-1}{x\left(x-4\right)}\)
=>\(\left(2x+8\right)\cdot\left(x^2-4x\right)=\left(x^2-4\right)\left(x+1\right)\)
=>2x^3-8x^2+8x^2-32x=x^3+x^2-4x-4
=>2x^3-32x=x^3-4x-4
=>x^3-28x+4=0
=>\(x\in\left\{-5,36;5,22;0,14\right\}\)
