Làm 2 câu các câu còn lại tương tự!
a, \(E=-x^2+4x-5=-\left(x^2-4x+5\right)\)
\(=-\left(x^2-2x-2x+4+1\right)=-\left[\left(x-2\right)^2+1\right]\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x-2\right)^2+1\ge1\Rightarrow-\left[\left(x-2\right)^2+1\right]\le-1\)
Hay \(E\le-1\) với mọi giá trị của \(x\in R\).
Để \(E=-1\) thì \(-\left[\left(x-2\right)^2+1\right]=-1\)
\(\Rightarrow\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy.............
b, \(F=-2x^2+2x-1=-\left(2x^2-2x+1\right)\)
\(=-\left(2x^2-x-x+\dfrac{1}{2}-\dfrac{3}{2}\right)\)
\(=-\left[\left(2x-1\right)^2-\dfrac{3}{2}\right]\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(2x-1\right)^2-\dfrac{3}{2}\ge-\dfrac{3}{2}\Rightarrow-\left[\left(2x-1\right)^2-\dfrac{3}{2}\right]\le\dfrac{3}{2}\)
Hay \(F\le\dfrac{3}{2}\) với mọi giá trị của \(x\in R\).
Để \(F=\dfrac{3}{2}\) thì \(-\left[\left(2x-1\right)^2-\dfrac{3}{2}\right]=\dfrac{3}{2}\)
\(\Rightarrow\left(2x-1\right)^2=0\Rightarrow x=\dfrac{1}{2}\)
Vậy.............
7, \(G=-4x^2+12x-7\)
\(=-4\left(x^2-3x+\dfrac{7}{4}\right)\)
\(=-4\left(x^2-\dfrac{3}{2}.x.2+\dfrac{9}{4}-\dfrac{2}{4}\right)\)
\(=-4\left(x-\dfrac{3}{2}\right)^2+2\le2\)
Dấu " = " khi \(-4\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(MAX_G=2\) khi \(x=\dfrac{3}{2}\)
8, \(H=-2x^2+4x-15\)
\(=-2\left(x^2-2x+\dfrac{15}{2}\right)\)
\(=-2\left(x^2-2x+1+\dfrac{13}{2}\right)\)
\(=-2\left(x-1\right)^2-13\le-13\)
Dấu " = " khi \(-2\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy \(MAX_H=-13\) khi x = 1
9, \(K=-x^4+2x^2-2\)
\(=-\left(x^2-2x^2+1+1\right)\)
\(=-\left(x^2-1\right)^2-1\le-1\)
Dấu " = " khi \(-\left(x^2-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy \(MAX_K=-1\) khi \(x=\pm1\)
10, \(J=-3x^2+15x-9\)
\(=-3\left(x^2-\dfrac{5}{2}.x.2+\dfrac{10}{4}+\dfrac{2}{4}\right)\)
\(=-3\left(x-\dfrac{5}{2}\right)^2-\dfrac{3}{2}\le\dfrac{-3}{2}\)
Dấu " = " khi \(-3\left(x-\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(MAX_J=\dfrac{-3}{2}\) khi \(x=\dfrac{5}{2}\)
