a)\(A=2x+1-x^2=2-\left(x^2-2x+1\right)=2-\left(x-1\right)^2\le2;\forall x\)
\(\Rightarrow A_{max}=2\Leftrightarrow x=1\)
b)\(B=4x-4x^2-5=-4-\left(4x^2-4x+1\right)=-4-\left(2x-1\right)^2\le-4;\forall x\)
\(\Rightarrow B_{max}=-4\Leftrightarrow x=\dfrac{1}{2}\)
a) `A=2x+1-x^2`
`=-(x^2-2x-1)`
`=-(x^2-2x+1)+2`
`=-(x-1)^2+2`
Có: `-(x-1)^2 <= forall x => -(x-1)^2+2 <=2`
`=> A_(max)=2 <=> x=1`
b) `B=4x-4x^2-5`
`=-(4x^2-4x+5)`
`=-(4x^2-4x+1)-4`
`=-[(2x)^2-2.2x.1+1^2]-4`
`=-(2x-1)^2+4`
`=> B_(max)=4 <=> x=1/2`
a) Ta có: \(A=-x^2+2x+1\)
\(=-\left(x^2+2x-1\right)\)
\(=-\left(x+1\right)^2+2\le2\forall x\)
Dấu '=' xảy ra khi x=-1
b) Ta có: \(B=-4x^2+4x-5\)
\(=-\left(4x^2-4x+5\right)\)
\(=-\left(2x-1\right)^2-4\le-4\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
