a) Đặt \(A=10+2x-5x^2\)

\(-A=5x^2-2x-10\)

\(-5A=25x^2-10x-50\)

\(-5A=\left(25x^2-10x+1\right)-51\)

\(-5A=\left(5x-1\right)^2-51\)

Do \(\left(5x-1\right)^2\ge0\forall x\)

\(\Rightarrow-5A\ge-51\)

\(A\le\frac{51}{5}\)

Dấu "=" xảy ra khi : \(5x-1=0\Leftrightarrow x=\frac{1}{5}\)

Vậy Max A = \(\frac{51}{5}\Leftrightarrow x=\frac{1}{5}\)

b) Đặt \(B=x^2-6x+10\)

\(B=\left(x^2-6x+9\right)+1\)

\(B=\left(x-3\right)^2+1\)

Mà \(\left(x-3\right)^2\ge0\forall x\)

\(B\ge1\)

Dấu "=" xảy ra khi :

\(x-3=0\Leftrightarrow x=3\)

Vậy Min B \(=1\Leftrightarrow x=3\)

1: (5x+3)^2>=0

=>2(5x+3)^2>=0

=>A<=6

Dấu = xảy ra khi x=-3/5

2: (x+9)^2+10>=10 

=>B<=13/10

Dấu = xảy ra khi x=-9

3: -3(2x-1)^2<=0

=>-3(2x-1)^2-7<=-7

Dấu = xảy ra khi x=1/2

Ta có: \(B=-\left(2x^2-5x+8\right)\)

 \(\Rightarrow B=-\left[2x^2-2.2x.\frac{5}{4}+\left(\frac{5}{4}\right)^2\right]+\frac{27}{4}\)

\(\Rightarrow B=-\left(2x-\frac{5}{4}\right)^2+\frac{27}{4}\)

\(\Rightarrow B=27-\left(2x-\frac{5}{4}\right)^2\)

Vì \(\left(2x-\frac{5}{4}\right)^2\ge0\Rightarrow B\le\frac{27}{4}\)

Dấu "=" xảy ra khi \(2x-\frac{5}{4}=0\Rightarrow x=\frac{5}{8}\)

Vậy B_max=\(\frac{27}{4}\) khi \(x=\frac{5}{8}\)

-B = 2x^2 - 5x + 8 = 2.(x^2 - 5/2 x + 25/16 ) + 39/8 = 2.(x-5/4)^2 + 39/8 >= 39/8

=> B <= -39/8

Dấu "=" xảy ra <=> x-5/4 = 0 <=> x=5/4

Vậy Max B = -39/8 <=> x=5/4

mình làm cho nhé :

-2x²+5x-8 

=-(2x²-5x)-8

= -2(x²-2.\(\frac{5}{2}\).x +(\(\frac{5}{2}\))² - (\(\frac{5}{2}\))²) -8

-2(x-\(\frac{5}{2}\))²-\(\frac{9}{2}\)

Nhận xét : -2(x-\(\frac{5}{2}\))²  <hoạc bằng 0 ; -2(x-\(\frac{5}{2}\))²  -\(\frac{9}{2}\)>hoặc bàng 0

=>B_(min)=\(\frac{9}{2}\)dấu = xảy ra khi x =\(\frac{5}{2}\)công sức của 

1/

a, \(A=4x^2-4x+5=4x^2-4x+1+4=\left(2x-1\right)^2+4\ge4\)

Dấu "=" xảy ra khi x=1/2

Vậy Amin=4 khi x=1/2

b, \(B=3x^2+6x-1=3\left(x^2+2x+1\right)-4=3\left(x+1\right)^2-4\ge-4\)

Dấu "=" xảy ra khi x=-1

Vậy Bmin = -4 khi x=-1

2/

a, \(A=10+6x-x^2=-\left(x^2-6x+9\right)+19=-\left(x-3\right)^2+19\le19\)

Dấu "=" xảy ra khi x=3

Vậy Amax = 19 khi x=3

b, \(B=7-5x-2x^2=-2\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)+\frac{31}{8}=-2\left(x-\frac{5}{4}\right)^2+\frac{31}{8}\le\frac{31}{8}\)

Dấu "=" xảy ra khi x=5/4

Vậy Bmax = 31/8 khi x=5/4

2) \(P=\frac{4}{2x^2+2xy+y^2+5x+20}=\frac{4}{\left(x^2+2xy+y^2\right)+\left(x^2+5x+\frac{25}{4}\right)+\frac{75}{4}}\)

\(=\frac{4}{\left(x+y\right)^2+\left(x+\frac{5}{2}\right)^2+\frac{75}{4}}\)

Để P đạt GTLN 

=> Mẫu thức đạt GTNN

mà \(\left(x+y\right)^2+\left(x+\frac{5}{2}\right)^2+\frac{75}{4}\ge\frac{75}{4}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y=0\\x+\frac{5}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{5}{2}\\y=\frac{5}{2}\end{cases}}\)

Thay x = -5/2 và y = 5/2 vào P 

Khi đó P = \(\frac{4}{\left(-\frac{5}{2}+\frac{5}{2}\right)^2+\left(-\frac{5}{2}+\frac{5}{2}\right)^2+\frac{75}{4}}=\frac{4}{\frac{75}{4}}=\frac{16}{75}\)

Vậy Max P = 16/75 <=> x = -5/2 ; y = 5/2

1) Ta có P = x² + 2xy + 3y² + 5y + 10

= (x² + 2xy + y²) + (2y² + 5y + 10) 

= \(\left(x+y\right)^2+2\left(y^2+\frac{5}{2}y+5\right)=\left(x+y\right)^2+2\left(y^2+\frac{5}{2}y+\frac{25}{16}+\frac{55}{16}\right)\)

= \(\left(x+y\right)^2+2\left(y+\frac{5}{4}\right)^2+\frac{55}{8}\ge\frac{55}{8}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y=0\\y+\frac{5}{4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{4}\\y=-\frac{5}{4}\end{cases}}\)

Vạy Min P = 55/8 <=> x = 5/4 ; y = -5/4 

Max A = -7

sory lộn 

TT

\(A=-2x^2+8x-15\)

\(-A=2x^2-8x+15\)

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

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

Mà  \(\left(x-2\right)^2\ge0\forall x\Rightarrow2\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge7\)

\(\Leftrightarrow A\le-7\)

Dấu "=" xảy ra khi : 

\(x-2=0\Leftrightarrow x=2\)

Vậy  \(A_{Max}=7\Leftrightarrow x=2\)

\(B=-5x\left(x+2\right)\)

\(B=-5x^2-10x\)

\(-B=5x^2+10x\)

\(-B=5\left(x^2+2x+1\right)-5\)

\(-B=5\left(x+1\right)^2-5\)

Mà  \(\left(x+1\right)^2\ge0\forall x\Leftrightarrow5\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow-B\ge-5\)

\(\Leftrightarrow B\le5\)

Dấu "=" xảy ra khi : 

\(x+1=0\Leftrightarrow x=-1\)

Vậy  \(B_{Max}=5\Leftrightarrow x=-1\)

a) ta có A = (2x-1)²+ ( x+2)= 4x²- 4x +1 +x+2= 4x² -3x +3 = 4x²-2*2x* \(\frac{3}{4}\)+ \(\frac{9}{16}\)+ \(\frac{39}{16}\)

= (2x-\(\frac{3}{4}\))²+ \(\frac{39}{16}\)

=> (2x-\(\frac{3}{4}\))²>=0

=> A >= \(\frac{39}{16}\)

dấu = sảy ra khi x=\(\frac{3}{2}\)

vậy A_{(min) = \(\frac{39}{16}\) khi x=\(\frac{3}{2}\)}

b) lm tương tự B_(min)= -\(\frac{25}{4}\) khi x= \(\frac{5}{2}\)

c) đặt dấu trừ ra ngoài vậy C_(max)=0 khi x=2

1) \(P=-2x^2-12x=-2\left(x^2+6x+9\right)+18=-2\left(x+3\right)^2+18\le18\)

\(maxP=18\Leftrightarrow x=-3\)

2) \(Q=-5x^2+10x=-5\left(x^2-2x+1\right)+5=-5\left(x-1\right)^2+5\le5\)

\(maxQ=5\Leftrightarrow x=1\)

3) \(A=-3x^2+12x-6=-3\left(x^2-4x+4\right)+6=-3\left(x-2\right)^2+6\le6\)

\(maxA=6\Leftrightarrow x=2\)

4) \(B=-2x^2-24x+12=-2\left(x^2+12x+36\right)+84=-2\left(x+6\right)^2+84\le84\)

\(maxB=84\Leftrightarrow x=-6\)