\(M=\frac{\left(x^2+5\right)-\left(x^2-4x+4\right)}{x^2+5}=1-\frac{\left(x-2\right)^2}{x^2+5}\le1\)

Dấu = xảy ra khi x - 2 =0 <=> x = 2

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

Bài 5:

a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)

\(minA=5\Leftrightarrow x=2\)

b) \(B=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)

Bài 4:

a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

\(maxM=7\Leftrightarrow x=2\)

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

\(maxN=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(P=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)

\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)

2−2x+5

2+2x−5)

2+2x+1)+6

2+6

2≤0∀x

2+6≤6∀x

Dấu "=" xảy ra ⇔

$x=-1$

Vậy $MA{X}_M=6\iff x=-1$

2+3

22−4x−3)

2−4x+4−7)

2−7]

2+7

22+7≤7

2=0⇔x=2

Vậy MAXA=7 khi x = 2

mới chỉ có GTLN thôi XD

\(M=4x-x^2-5\)

\(-M=x^2-4x+5\)

\(-M=x^2-2\cdot2\cdot x+2^2+1\)

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

\(\left(x-2\right)^2\ge0\Rightarrow\left(x-2\right)^2+1\ge1\)

\(\Rightarrow-M\ge1\)

\(\Rightarrow M\le1\)

dấu "=" xảy ra khi : 

\(\left(x-2\right)^2=0\Rightarrow x-2=0\Rightarrow x=2\) 

M = -x² - y² + 4x - 2y + 5

<=> M = - x² + 4x - 4 - y² - 2y - 1 + 10

<=> M = -(x² - 4x+ 4) - (y² + 2y + 1) + 10

<=> M = -(x - 2)² - (y + 1)² + 10

Do: (x - 2)² lớn hơn hoặc bằng 0 <=> -(x - 2)² bé hơn hoặc bằng 0

(y + 1)² lớn hơn hoặc bằng 0 <=> -(y + 1)² bé hơn hoặc bằng 0

=> M bé hơn hoặc bằng 10

Dấu "=" xảy ra khi: -(x - 2)² = 0 và -(y + 1)² = 0

<=> x = 2 và y = -1

Vậy GTNN của M là 10 khi và chỉ khi x = 2 và y = -1 

\(M=-\left(x^2-4x+16\right)-\left(y^2+2y+1\right)+20\)

\(=-\left(x-4\right)^2-\left(y+1\right)^2+20\le20\)

\(M_{max}=20\)dấu bằng sảy ra khi \(\hept{\begin{cases}x-4=0\\y+1=0\end{cases}}=\orbr{\begin{cases}x=4\\y=-1\end{cases}}\)

M = \(-\left(x^2+y^2-4x+2y-5\right)\)

M = \(-\left(x^2-4x+4+y^2+2y+1\right)+10\)

M = \(-\left[\left(x-2\right)^2+\left(y+1\right)^2\right]+10\le10\) với mọi x, y.

Đẳng thức xảy ra \(\hept{\begin{cases}x-2=0\Leftrightarrow x=2\\y+1=0\Leftrightarrow y=-1\end{cases}}\)

Vậy GTLN của M là 10 khi x = 2, y = -1.

1:

a: =x^2-7x+49/4-5/4

=(x-7/2)^2-5/4>=-5/4

Dấu = xảy ra khi x=7/2

b: =x^2+x+1/4-13/4

=(x+1/2)^2-13/4>=-13/4

Dấu = xảy ra khi x=-1/2

e: =x^2-x+1/4+3/4=(x-1/2)^2+3/4>=3/4

Dấu = xảy ra khi x=1/2

f: x^2-4x+7

=x^2-4x+4+3

=(x-2)^2+3>=3

Dấu = xảy ra khi x=2

2:

a: A=2x^2+4x+9

=2x^2+4x+2+7

=2(x^2+2x+1)+7

=2(x+1)^2+7>=7

Dấu = xảy ra khi x=-1

b: x^2+2x+4

=x^2+2x+1+3

=(x+1)^2+3>=3

Dấu = xảy ra khi x=-1

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

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

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