\(A=-\left(x^2-4x-3\right)=-\left(x^2-4x+4-7\right)=7-\left(x-2\right)^2\le7\Rightarrow A_{max}=7\Leftrightarrow x-2=0\Rightarrow x=2\)
mk tra loi cau b con lai bn dua vao de giai nhé
b. x - x^2 = -(x^2 - x)
= -[ (x^2 - 2.x.1/2 +(1/2)^2-(1/2)^2
= -[(x-1/2)^2 - (1/2)^2]
= -(x-1/2)^2 + 1/4 = 1/4 - (x-1/2)^2
Vì (x-1/2)^2 >=0 nên 1/4 - (x-1/2)^2 <=1/4 với mọi x
Do đó đa thức đã cho có gtln la 1/4 tại x = 1/2
( ý 2 là thêm bớt hạng tử nha)
\(B=-\left(x^2-x\right)=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}\right)=\frac{1}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{1}{4}\Rightarrow B_{max}=\frac{1}{4}\Leftrightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
\(C=-\left(2x^2-2x+5\right)=-2\left(x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{5}{2}-\frac{1}{4}\right)=-2\left(x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{9}{4}\right)\)
\(=-\left(x-\frac{1}{2}\right)^2-\frac{9}{2}^2\le-\frac{9}{2}\)
\(C_{max}=-\frac{9}{2}\Leftrightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
