\(M=x+\sqrt{x+\dfrac{1}{4}+2.\sqrt{x+\dfrac{1}{4}}.\dfrac{1}{2}+\dfrac{1}{4}}\)
\(M=x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=x+\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\text{}\text{}\) (do căn + 1 số luôn dương)
\(M=x+\dfrac{1}{4}+2.\sqrt{x+\dfrac{1}{4}}.\dfrac{1}{2}+\dfrac{1}{4}\)
\(M=\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2\)
ta thấy \(\sqrt{x+\dfrac{1}{4}}\ge0\)
\(\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2\ge\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
Dấu = xảy ra khi \(x+\dfrac{1}{4}=0\Leftrightarrow x=-\dfrac{1}{4}\)
vậy \(M_{min}=\dfrac{1}{4}\) khi \(x=-\dfrac{1}{4}\)
b) ta có \(M=\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2\)
ĐK \(x\ge\dfrac{-1}{4}\)
\(\Leftrightarrow\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=9\)
\(\Leftrightarrow\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2-9=0\)
\(\Leftrightarrow\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}-3\right)\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+\dfrac{1}{4}}=\dfrac{5}{2}\\\sqrt{x+\dfrac{1}{4}}=\dfrac{-7}{2}\left(vl\right)\end{matrix}\right.\)
\(\Leftrightarrow x+\dfrac{1}{4}=\dfrac{25}{4}\Rightarrow x=6\)
