\(A=x^2+2x^4+3\)
\(=2\left(x^4+\frac{1}{2}x^2+\frac{3}{2}\right)\)
\(=2\left(x^4+2x^2.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}+\frac{3}{2}\right)\)
\(=2\left[\left(x^2+\frac{1}{4}\right)^2+\frac{23}{16}\right]\)
\(=2\left(x^2+\frac{1}{4}\right)^2+\frac{23}{8}\)
\(A_{min}\Leftrightarrow\left(x^2+\frac{1}{4}\right)_{min}\Leftrightarrow x^2_{min}\Rightarrow x=0\)
Với \(x=0\Rightarrow A=0^2+2.0^4+3=3\)
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ta có:
\(x^2\ge0\forall x\in R\)
\(2x^4\ge0\forall x\in R\)
=>A\(\ge3\)
=>Amin =3
vậy.......
hc tốt
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