Bài 1:
a: Ta có: \(\dfrac{x-2}{5}-1=\dfrac{2x-1}{2}\)
\(\Leftrightarrow2x-4-10=10x-5\)
\(\Leftrightarrow-8x=9\)
hay \(x=-\dfrac{9}{8}\)
b: Ta có: \(\dfrac{3}{5-x}=\dfrac{2}{25-x^2}\)
\(\Leftrightarrow3\left(5+x\right)=2\)
\(\Leftrightarrow x=-13\)
c: Ta có: \(x\left(2-x^2\right)+\left(x-2\right)\left(x^2+2x+4\right)>3\)
\(\Leftrightarrow2x-x^3+x^3-8>3\)
\(\Leftrightarrow2x>11\)
hay \(x>\dfrac{11}{2}\)