Question

Chứng tỏ:
A=x²+4x+5>0
Với mọi x thuộc R

Answer 1

\(\Rightarrow A=x^2+4x+4+1=\left(x+2\right)^2+1\ge1>0\forall x\in R\)

Answer 2

\(A=x^2+4x+5\)

\(=x^2+4x+4+1\)

\(=(x+2)^2+1\)

Vì \(x+2)^2 \ge 0\)

\(\Rightarrow (x+2)^2+1 \ge 1>0 \forall x \in R\)

Hay \(A=x^2+4x+5>0 \forall x \in R\)

Answer 3

\(A=x^2+4x+5=x^2+4x+4+1=\left(x+2\right)^2+1\)

vì \(\left(x+2\right)^2\ge0=>\left(x+2\right)^2+1\ge1\left(\forall x\in R\right)\)

\(=>\left(x+2\right)^2+1>0\left(\forall x\in R\right)\)

Answer 4

Ta có: \(A=x^2+4x+5\)

\(=x^2+4x+4+1\)

\(=\left(x+2\right)^2+1>0\forall x\)