Question

Giải bất phương trình

x^2>= 1

x^2 < 1

x^2+3x>=0

x^2+3x+3 >=0

Answer 1

a, \(x^2\)≥1

\(\Leftrightarrow\) x>1

b, \(x^2\)<1

\(\Rightarrow\) x∈∅

c, \(x^2\)+3x ≥ 0

\(\Leftrightarrow\) \(x^2\)≥-3x

\(\Leftrightarrow\) x≥-3

d, \(x^2\)+3x+3≥0

\(\Leftrightarrow\) \(\left(x+\dfrac{3}{2}\right)^2\)+\(\dfrac{3}{4}\)≥0+\(\dfrac{3}{4}\)

\(\Leftrightarrow\) \(x^2\)+\(\dfrac{3}{2}^2\)≥0

\(\Leftrightarrow\)\(x^2\)≥\(\dfrac{9}{4}\)

\(\Leftrightarrow\)x≥\(\dfrac{3}{2}\)