a, \(A=x^2-6x+11\)

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

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

Ta có :

\(\left(x-3\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x-3\right)^2+2\ge2\) với mọi x

Dấu = xảy ra \(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)

Vậy \(Min_A=2\Leftrightarrow x=3\)

b, \(B=2x^2+10x-1\)

\(=2\left(x^2+5x+\dfrac{25}{4}\right)-\dfrac{25}{2}-1\)

\(=2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\)

Lập luận tương tự câu a

c, \(C=5x-x^2\)

\(=-\left(x^2-5x+\dfrac{25}{2}\right)+\dfrac{25}{2}\)

\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{2}\)

Lập luận tương tự câu a

A = 0

B = 0, = 1/5

k nha

\(A=2x^2+10x-1=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge-\frac{27}{2}\)

=> Min A \(=-\frac{27}{2}\Leftrightarrow x=-\frac{5}{2}\)

\(B=5x^2-x=5\left(x-\frac{1}{10}\right)^2-\frac{1}{20}\ge-\frac{1}{20}\)

=> Min B \(=-\frac{1}{20}\Leftrightarrow x=\frac{1}{10}\)

\(A=x^2-4x+7=\left(x^2-4x+4\right)+3=\left(x-2\right)^2+3\)

Vì: \(\left(x-2\right)^2\ge0\)

=> \(\left(x-2\right)^2+3\ge3\)

Vậy GTNN của A là 3 khi x=2

\(B=2x^2+12x-1=2\left(x^2+6x+9\right)-19=2\left(x+3\right)^2-19\)

Vì: \(2\left(x+3\right)^2\ge0\)

=> \(2\left(x+3\right)^2-19\ge-19\)

Vậy GTNN của B là -19 khi x=-3

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

Vì: \(-\left(x-\frac{5}{2}\right)^2\le0\)

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

Vậy GTLN của C là \(\frac{25}{4}\) khi \(x=\frac{5}{2}\)

\(A=x^2-4x+1\)
\(A=x^2-4x+4-3\)
\(A=\left(x-2\right)^2-3\)
Min A = -3
Min A xảy ra khi (x-2)²=0
                           x-2=0
                           x=2
 

A đến C là tìm GTNN

\(A=x^2-4x+1=\left(x-2\right)^2-3\ge-3\)

Dấu "=" xảy ra ⇔ x=2

\(B=2x^2-x+1=2\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)+\dfrac{7}{8}=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{4}\)

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

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

D đến F là tìm GTLN

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

Do (x-1)²≥0 ⇔-(x-1)²≤0 ⇔ D≤-1

Dấu "=" xảy ra ⇔ x=1

\(D=-x^2+x-3=-\left(x^2-2.\dfrac{1}{2}+\dfrac{1}{4}\right)-\dfrac{11}{4}=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)

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

\(F=-3x^2+x-2=-3\left(x^2-2.\dfrac{1}{6}+\dfrac{1}{36}\right)-\dfrac{23}{12}=-3\left(x-\dfrac{1}{6}\right)-\dfrac{23}{12}\le-\dfrac{23}{12}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{6}\)

Ta có : A = 2x² + 10x - 15 

= 2x² + 10x - \(\frac{50}{4}-\frac{5}{2}\)

= 2(x² + 5x - \(\frac{25}{4}\)) - \(\frac{5}{2}\)

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

Mà ; 2(x - \(\frac{5}{2}\) )²  \(\ge0\forall x\)

Nên : 2(x - \(\frac{5}{2}\) )² - \(\frac{5}{2}\) \(\ge-\frac{5}{2}\forall x\)

Vậy A_min = \(-\frac{5}{2}\) , dấu bằng xảy ra khi x = \(\frac{5}{2}\)

\(A=x^2-4x+1=\left(x^2-4x+4\right)-3=\left(x-2\right)^2-3\ge-3\)

Vậy \(A_{Min}=-3khix=2\)

giải câu b trc nha

= ((x-1)^2+2009]/x^2=(x-1)^2/x^2+2009

vậy min=2009 khi x=1

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