\(A=\left(x^2+2x\frac{3}{2}+\frac{9}{4}+\frac{7}{4}\right)^2\)
\(\Leftrightarrow A=\left[\left(x+\frac{3}{2}\right)^2+\frac{7}{4}\right]^2\)
Vì \(\left(x+\frac{3}{2}\right)^2\ge0\) với \(\forall x\in R\)
\(\Rightarrow\left(x+\frac{3}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)với \(\forall x\in R\)
\(\Rightarrow A=\left[\left(x+\frac{3}{2}\right)^2+\frac{7}{4}\right]^2\ge\frac{49}{16}\)với \(\forall x\in R\)
Vậy GTNN của \(A=\frac{49}{16}\) .Dấu "=" xảy ra khi \(x+\frac{3}{2}=0\)
\(\Leftrightarrow x=-\frac{3}{2}\)
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