\(a,\) \(A=-x^2+2x+3\) \(=-\left(x^2-2x+1\right)+4\)\(=-\left(x-1\right)^2+4\)
Mà \(-\left(x-1\right)^2\le0\left(\forall x\right)\)
\(\Rightarrow-\left(x-1\right)^2+4\le4\left(\forall x\right)\)
Vậy MaxA = 4 \(\Leftrightarrow\)x - 1 = 0\(\Leftrightarrow x=1\)
b,\(B=-10x-x^2+1\)
=\(-\left(x^2+10x+25\right)+26\)=\(-\left(x+5\right)^2+26\)
Mà \(-\left(x+5\right)^2\le0\left(\forall x\right)\)
\(\Rightarrow-\left(x+5\right)^2+26\le26\)
Vậy Max B = 26\(\Leftrightarrow x+5=0\Leftrightarrow x=-5\)
c, \(C=x-x^2=-\left(x^2-x\right)\)
\(=-\left(x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{1}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)
Mà \(-\left(x-\dfrac{1}{2}\right)^2\le0\left(\forall x\right)\)
\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\left(\forall x\right)\)
Vậy Max A = \(\dfrac{1}{4}\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
d, D = \(-\left(3x^2-3x+8\right)\)=\(-3\left(x^2-x+\dfrac{8}{3}\right)\)=\(-3\left(x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}\right)-7\dfrac{1}{4}\)
=\(-3\left(x-\dfrac{1}{2}\right)^2-7\dfrac{1}{4}\)
Mà \(-3\left(x-\dfrac{1}{2}\right)^2\le0\left(\forall x\right)\)
\(\Rightarrow-3\left(x-\dfrac{1}{2}\right)^2-7\dfrac{1}{4}\le-7\dfrac{1}{4}\left(\forall x\right)\)
Vậy MaxA=\(-7\dfrac{1}{4}\)\(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
