\(\dfrac{x-1}{x+2}=\dfrac{4}{5}=>\left(x-1\right).5=\left(x+2\right).4\)
=> \(5x-5=4x+8\)
=> \(5x-4x=8+5\)
=> x = 13
\(x-\dfrac{1}{x}+2=\dfrac{4}{5}\)
<=> \(\dfrac{x^2}{x}-\dfrac{1}{x}=-\dfrac{6}{5}\)
<=> \(\dfrac{x^2}{x}-\dfrac{1}{x}+\dfrac{6}{5}=0\)
<=> \(\dfrac{5x^2}{5x}-\dfrac{5}{5x}+\dfrac{6x}{5x}=0\)
<=> \(5x^2+6x-5=0\)
nó nhầm nhầm:v