\(x-x^2-1\\ =-\left(x^2-x+1\right)\\ =-\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =-\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\\ \left(x-\dfrac{1}{2}\right)^2\ge0\forall x\in R\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\in R\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\in R\\ \Rightarrow-\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]< 0\forall x\in R\\ \Leftrightarrow x-x^2-1< 0\forall x\in R\)
Vậy \(x-x^2-1< 0\forall x\in R\)
Ta có: \(x-x^2-1\)
\(=-\left(x^2-x+1\right)\)
\(=-\left(x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1\right)\)
\(=-\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\)
\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\)
Vì \(-\left(x-\dfrac{1}{2}\right)^2\le0\forall x\in R\)
\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le\dfrac{-3}{4}< 0\forall x\in R\)
-> ĐPCM.
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