a, \(x-x^2=-\left(x^2-x\right)=-\left(x^2-\dfrac{1}{2}x-\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}\right)\)
\(=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\right]\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(\Rightarrow-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\right]\le\dfrac{1}{4}\)
Để \(-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\right]=\dfrac{1}{4}\) thì \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Rightarrow x=\dfrac{1}{2}\)
Vậy...............
b, \(4x-x^2+1=-\left(x^2-4x-1\right)\)
\(=-\left(x^2-2x-2x+4-5\right)=-\left[\left(x-2\right)^2-5\right]\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x-2\right)^2-5\ge-5\)
\(\Rightarrow-\left[\left(x-2\right)^2-5\right]\le5\)
Để \(-\left[\left(x-2\right)^2-5\right]=5\) thì \(\left(x-2\right)^2=0\)
\(\Rightarrow x=2\)
Vậy...............
Chúc bạn học tốt!!!
a) \(x-x^2\)
\(=-\left(x^2-x\right)\)
\(=-\left(x^2-2.x.1+1^2-1\right)\)
\(=-\left[\left(x-1\right)^2-1\right]\)
\(=-\left(x-1\right)^2+1\)
Vì \(-\left(x-1\right)^2\le0\forall x\Rightarrow-\left(x-1\right)^2+1\le1\forall x\)
Dấu "=" xảy ra khi \(\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\)
Vậy \(GTLN=1\) khi \(x=1.\)
b) \(4x-x^2+1\)
\(=-\left(x^2-4x-1\right)\)
\(=-\left(x^2-2.x.2+2^2-4-1\right)\)
\(=-\left[\left(x-2\right)^2-5\right]\)
\(=-\left(x-2\right)^2+5\)
Do \(-\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-2\right)^2+5\le5\forall x\)
Dấu "=" xảy ra khi \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
Vậy \(GTLN=5\) khi \(x=2.\)
