Question

\(\dfrac{2x+x^2+1}{x^2+1}\le0\)

Answer 1

vì x2 + 1 > 0
x2 + 2x + 1 = (x + 1)2 >= 0
=> (x + 1)2 = 0
=> x = -1

:)

Answer 2

\(\dfrac{2x+x^2+1}{x^2+1}\le0\Leftrightarrow\dfrac{\left(x+1\right)^2}{x^2+1}\le0\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x+1=0\)  \(\Leftrightarrow x=-1\)