\(N=2x-2x^2-5\)
\(=-2\left(-x+x^2+\frac{5}{2}\right)\)
\(=-2\left(x^2-x+\frac{5}{2}\right)\)
\(=-2\left(x^2-2.\frac{1}{2}.x+\frac{1}{4}+\frac{9}{4}\right)\)
\(=-2\left[\left(x^2-2.\frac{1}{2}.x+\frac{1}{4}\right)+\frac{9}{4}\right]\)
\(=-2\left[\left(x-\frac{1}{2}\right)^2+\frac{9}{4}\right]\)
\(=-2.\left(x-\frac{1}{2}\right)^2-2.\frac{9}{4}\)
\(=-2.\left(x-\frac{1}{2}\right)^2-\frac{9}{2}\le-\frac{9}{2}\)
Dấu = xảy ra khi:
\(-2\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(\Rightarrow x=0+\frac{1}{2}=\frac{1}{2}\)
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