Question

Giải bất phương trình: \(\dfrac{x^2+2x+2}{x+1}>\dfrac{x^2+4x+5}{x+2}-1\)

Answer 1

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x^2+2x+2\right)-\left(x^2+4x+5\right)\left(x+1\right)+\left(x^2+3x+2\right)}{\left(x+1\right)\left(x+2\right)}>0\)

\(\Leftrightarrow\dfrac{x^3+4x^2+6x+4-\left(x^3+5x^2+9x+5\right)+x^2+3x+2}{\left(x+1\right)\left(x+2\right)}>0\)

=>\(\dfrac{x^3+5x^2+9x+6-x^3-5x^2-9x-5}{\left(x+1\right)\left(x+2\right)}>0\)

=>(x+1)(x+2)>0

=>x>-1 hoặc x<-2