\(x^2+3x+1\)
=\(\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{5}{4}\)
=\(\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\)
Ta có:\(\left(x+\dfrac{3}{2}\right)^2\ge0\) Với mọi x
=>\(\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\)
Dấu "=" xảy ra <=>\(\left(x+\dfrac{3}{2}\right)^2=0\)
<=>\(x+\dfrac{3}{2}=0\)
<=>\(x=\dfrac{-3}{2}\)
\(\dfrac{x^2+3}{x+1}=\dfrac{x^2-1+4}{x+1}=x-1+\dfrac{4}{x+1}\)
\(=x+1+\dfrac{4}{x+1}-2\ge2\cdot\sqrt{4}-2=2\)
Dấu '=' xảy ra khi \(\left(x+1\right)^2=4\)
=>x+1=2 hoặc x+1=-2
=>x=1 hoặc x=-3
