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

a, (x-1)(x-3)+11

=x²-3x-x+3+11

=(x-2)²+10

Vì..................................

b,5-4x²+4x

=-(4x²-4x+4)+9

=-(2x-2)²+9

...........................................................

A = - 4\(x\)² + 5\(x\) - 3

A = -( 4\(x^2\) - 5\(x\) + \(\dfrac{25}{16}\))  - \(\dfrac{23}{16}\)

A = -( 2\(x\) - \(\dfrac{5}{4}\))² - \(\dfrac{23}{16}\)

Vì ( 2\(x\) - \(\dfrac{5}{4}\))² ≥ 0; ⇒ - ( 2\(x\) - \(\dfrac{5}{4}\))² ≤ 0 ⇒ -( 2 \(x\) - \(\dfrac{5}{4}\))² - \(\dfrac{23}{16}\) ≤ - \(\dfrac{23}{16}\)

A(max) = - \(\dfrac{23}{16}\) ⇔ 2\(x\) - \(\dfrac{5}{4}\) = 0 ⇔ \(x\) = \(\dfrac{5}{4}\): 2 = \(\dfrac{5}{8}\)

Kết luận giá trị lớn nhất của biểu thức là - \(\dfrac{23}{16}\) xáy ra khi \(x\) = \(\dfrac{5}{8}\)

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

\(A_{min}=-1\) khi \(x=-2\)

\(A=\dfrac{4x^2+4-4x^2+4x-1}{x^2+1}=4-\dfrac{\left(2x-1\right)^2}{x^2+1}\le4\)

\(A_{max}=4\) khi \(x=\dfrac{1}{2}\)

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

\(A_{min}=-1\) khi \(x=2\)

\(A=\dfrac{4x^2+4-4x^2-4x-1}{x^2+1}=4-\dfrac{\left(2x+1\right)^2}{x^2+1}\le4\)

\(A_{max}=4\) khi \(x=-\dfrac{1}{2}\)

`-4x^2+4x-3`

`=-4x^2+4x-1-2`

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

Vì `-(2x-1)^2 <= 0 AA x`

  `=>-(2x-1)^2-2 <= -2 AA x`

Hay `-4x^2 + 4x-3 <= -2 AA x`

  `=>Max =-2<=>x=1/2`

\(-4x^2+4x-3\)

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

\(=-\left(4x^2-4x+1\right)-2\)

\(=-\left(2x-1\right)^2-2\)

Ta thấy: \(\left(2x-1\right)^2\ge0\forall x\)

\(\Rightarrow-\left(2x-1\right)^2\le0\forall x\)

\(\Rightarrow-\left(2x-1\right)^2-2\le-2\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

giúp mình với càng nhanh càng tốt

ở đây không có vote 5 sao đâu bạn

a, Ta có: \(-\left(x-3\right)^2\le0,\forall x\)

 \(\Leftrightarrow-\left(x-3\right)^2-7\le-7\)

Vậy max A là \(-7\Leftrightarrow x=3\)

b, Ta có \(B=-x^2-2x-5=-\left(x+1\right)^2-4\le4\)

Dấu \("="\Leftrightarrow x=-1\)

c, Ta có \(C=-4x^2-4x+9=-\left(2x+1\right)^2+10\le10\)

Dấu \("="\Leftrightarrow x=-\dfrac{1}{2}\)

d, Ta có:\(D=-3y^2-6y+1\\ =-3\left(y^2+2y-\dfrac{1}{3}\right)=-3\left[\left(y+1\right)^2-\dfrac{4}{3}\right]=-3\left(y+1\right)^2+4\le4\)

Dấu \("="\Leftrightarrow y=-1\)