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 \(=1\Leftrightarrow x=3\)

có ai giúp em không

a) x² - 2x + 5 = (x - 1)² + 4 >= 4

Min là 4 khi x = 1

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

a) \(A=6x-x^2-11=-\left(x^2-6x+9\right)-2=-\left(x-3\right)^2-2\le-2\)

Dấu \(=\)khi \(x-3=0\Leftrightarrow x=3\).

b) \(B=x^2-5x-2=x^2-2.\frac{5}{2}x+\left(\frac{5}{2}\right)^2-\frac{33}{4}=\left(x-\frac{5}{2}\right)^2-\frac{33}{4}\ge-\frac{33}{44}\)

Dấu \(=\)khi \(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\).

\(A=\frac{2x^2+6x+10}{x^2+3x+3}=\frac{2\left(x^2+3x+3\right)+4}{x^2+3x+3}=2+\frac{4}{x^2+3x+3}\)

Để A đạt GTLN thì x²+3x+3 bé nhất

mà x²+3x+3=\(x^2+3.\frac{2}{3}x+\frac{2^2}{3^2}+\frac{23}{9}=\left(x+\frac{2}{3}\right)^2+\frac{23}{9}\ge\frac{23}{9}\)

Dấu "=" xảy ra khi \(x+\frac{2}{3}=0=>x=\frac{-2}{3}\)

lúc đó \(A=2+\frac{4}{\frac{23}{9}}=2+4.\frac{9}{23}=2+\frac{36}{23}=\frac{82}{23}\)

Vậy GTLN của \(A=\frac{82}{23}\)khi \(x=\frac{-2}{3}\)

a) \(6x-x^2-11\)

\(=-x^2+6x-11\)

\(=-\left(x^2-6x+11\right)\)

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

\(=-[\left(x-3\right)^2+2]\)

Mà: \(\left(x-3\right)^2\ge0\)

\(\Rightarrow-\left(x-3\right)^2\le0\)

\(\Rightarrow-\left(x-3\right)^2-2\le0-2\)

\(\Rightarrow A\le-2\)

Dấu '' = '' xảy ra khi: \(\left(x-3\right)^2=0\Rightarrow x=3\)

Vậy giá trị lớn nhất của biểu thức \(6x-x^2-11=-2\) khi \(x=3\)

b) \(x^2-5x-2\)

\(=\left(x^2-2.\frac{5}{2}x+\frac{25}{4}\right)-\frac{33}{4}\)

\(=\left(x-\frac{5}{2}\right)^2-\frac{33}{4}\)

Mà: \(\left(x-\frac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\frac{5}{2}\right)^2-\frac{33}{4}\ge\frac{-33}{4}\forall x\)

Dấu '' = '' xảy ra khi: \(x-\frac{5}{2}=0\Rightarrow x=\frac{5}{2}\)

Vậy giá trị nhỏ nhất của biểu thức \(x^2-5x-2=\frac{-33}{4}\)  khi \(x=\frac{5}{2}\)

`A=2x^2-2xy-6x+y^2+10`

`A=x^2-2xy+y^2+x^2-6x+10`

`A=(x-y)^2+x^2-6x+9+1`

`A=(x-y)^2+(x-3)^2+1`

Vì `(x-y)^2+(x-3)^2>=0=>A>=1`

Dấu "=" xảy ra khi `{(x-y=0),(x-3=0):}<=>x=y=3`

A=\(\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)+1=\left(x-y\right)^2+\left(x-3\right)^2+1\ge1\\ \)

dấu= xảy ra khi x=y=3

tick mik nha

a) \(A=x^2+6x+10\)

\(A=x^2+2\cdot x\cdot3+3^2+1\)

\(A=\left(x+3\right)^2+1\ge1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

b) \(B=2x^2+y^2+2xy+4x+15\)

\(B=\left(x^2+2xy+y^2\right)+\left(x^2+2\cdot x\cdot2+2^2\right)+11\)

\(B=\left(x+y\right)^2+\left(x+2\right)^2+11\ge11\forall x;y\)

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