ĐKXĐ:\(x\ge0\)
Đặt \(A=-x+\sqrt{x}-1+2\)
\(A=-x+\sqrt{x}+1\)
\(A=-x+2\sqrt{x}.\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{5}{4}\)
\(A=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\)
\(\forall x\in R\) ta có:\(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\)
\(\Rightarrow-\left(\sqrt{x}-\dfrac{1}{2}\right)^2\le0\)
\(\Rightarrow-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)
Hay \(A\le\dfrac{5}{4}\)
Dấu "=" xảy ra khi \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\)
\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{4}\left(TMĐK\right)\)
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