A=4x-x^2+3

= -x^2+4x+3

= -(x^2-4x-3)

= -(x^2-2*2x*1+1-4)

= -(x-1)^2+4 <4

GTLN của A là 4 khi x=1

Câu B có vấn đề bạn ơi

C=4x-x^2+1

= -x^2+4x+1

= -(x^2-4x-1)

= -(x^2-2*2x*1+1-2)

= -(x-1)^2+2 < 2

GTLN của C là 2 khi x=1

bạn xem lại đề bài 1 là GTNN hay GTLN nha

I zì:vv

a) Ta có: \(A=4x^2+4x+11=4x^2+4x+1=10=\left(2x+1\right)^2+10\ge10\forall x\)

Vậy Min_A=10 khi \(x=-\dfrac{1}{2}\)

b) Ta có: \(B=5-8x-x^2=-\left(x^2+8x-5\right)=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21\le21\forall x\)

Vậy Max_B=21 khi x=-4

a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)

\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)

\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)

Vậy Max_Q=10 khi x=2, y=-2

b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)

\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)

Vậy Max_A=14 khi x=-3

+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)

Vậy Max_B=5 khi x=-1/2, y=1/3

c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)

Vậy Min_P=2 khi x=1, y=-3

1) \(A=4x-x^2+3\)

\(A=-\left(x^2-4x-3\right)\)

\(A=-\left(x^2-4x+4\right)+7\)

\(A=-\left(x-2\right)^2+7\)

Mà: \(-\left(x-2\right)^2\le0\forall x\) nên: \(A=-\left(x-2\right)^2+7\le7\)

Dấu "=" xảy ra:

\(-\left(x-2\right)^2+7=7\)

\(\Rightarrow x=2\)

Vậy: \(A_{max}=7\) khi \(x=2\)

2) \(B=x-x^2\)

\(B=-x^2+x\)

\(B=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}\)

\(B=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)

Mà: \(-\left(x-\dfrac{1}{2}\right)^2\le0\forall x\) nên \(B=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

Dấu "=" xảy ra:
\(-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)

\(\Rightarrow x=\dfrac{1}{2}\)

Vậy: \(B_{max}=\dfrac{1}{4}\) với \(x=\dfrac{1}{2}\)