a, \(f\left(x\right)=\left(2-x\right)\left(x-1\right)\)
\(=2x-2-x^2+x\)
\(=-x^2+3x-2\)
\(=-\left(x^2-\dfrac{3}{2}x.2+\dfrac{9}{4}-\dfrac{1}{4}\right)\)
\(=-\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{1}{4}\right]\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
Dấu " = " xảy ra khi \(-\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(MAX_{f\left(x\right)}=\dfrac{1}{4}\) khi \(x=\dfrac{3}{2}\)
b, tương tự
c, \(h\left(x\right)=2x-3-x^2\)
\(=-\left(x^2-2x+3\right)\)
\(=-\left(x^2-2x+1+2\right)\)
\(=-\left[\left(x-1\right)^2+2\right]\)
\(=-\left(x-1\right)^2-2\le-2\)
Dấu " = " khi \(-\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy \(MAX_{h\left(x\right)}=-2\) khi x = 1
\(f\left(x\right)=\left(2-x\right)\left(x-1\right)\)
\(=-x^2+3x-2\)
\(=-\left(x^2-3x+2\right)\)
\(=-\left(x^2-\dfrac{3}{2}x-\dfrac{3}{2}x+2\right)\)
\(=-\left(x^2-\dfrac{3}{2}x-\dfrac{3}{2}x+\dfrac{9}{4}-\dfrac{1}{4}\right)\)
\(=-\left[x\left(x-\dfrac{3}{2}\right)-\dfrac{3}{2}\left(x-\dfrac{3}{2}\right)-\dfrac{1}{4}\right]\)
\(=-\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{1}{4}\right]\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{4}\)
Vì \(-\left(x-\dfrac{3}{2}\right)^2\le0\Rightarrow-\left(x-\dfrac{3}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(\Rightarrow f\left(x\right)\le\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(-\left(x-\dfrac{3}{2}\right)^2=0\)
\(\Rightarrow x=\dfrac{3}{2}.\)
Vậy \(Max_{f\left(x\right)}=\dfrac{1}{4}\) khi \(x=\dfrac{3}{2}.\)
Mấy câu kia tương tự.
