\(A=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
\(A_{max}=7\) khi \(x=2\)
\(B=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(B_{max}=\dfrac{1}{4}\) khi \(x=\dfrac{1}{2}\)
`A=4x-x^2+3`
`A=-(x^2-4x)+3`
`A=-(x^2-2x-2x+4)+4+3`
`A=-(x-2)^2+7<=7`
Dấu "=" xảy ra khi `x-2=0<=>x=2.`
`B=x-x^2`
`=-(x^2-x)`
`=-(x^2-x+1/4)+1/4`
`=-(x-1/2)^2+1/4<=1/4`
Dấu "=" xảy ra khi `x-1/2=0<=>x=1/2`
a) Ta có: \(A=-x^2+4x+3\)
\(=-\left(x^2-4x-3\right)\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left(x-2\right)^2+7\le7\forall x\)
Dấu '=' xảy ra khi x=2
b) Ta có: \(B=-x^2+x\)
\(=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
