1, \(3x^2-5x+4\)

\(=3\left(x^2-\frac{5}{3}x\right)+1=3\left(x^2-2.\frac{5}{6}x+\frac{25}{36}\right)+\frac{23}{12}=3\left(x-\frac{5}{6}\right)^2+\frac{23}{12}\)

Ta có: \(3\left(x-\frac{5}{6}\right)^2\ge0\forall x\Leftrightarrow3\left(x-\frac{5}{6}\right)^2+\frac{23}{12}\ge\frac{23}{12}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-\frac{5}{6}\right)^2=0\Leftrightarrow x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)

Vậy minA = \(\frac{23}{12}\Leftrightarrow x=\frac{5}{6}\)

2, Bạn thử kiểm tra lại đề bài xem

\(A=-2x^2+5x-8=-2\left(x^2-\frac{5}{2}x+4\right)\)

\(=-2\left(x^2-\frac{5}{2}x+\frac{25}{16}+\frac{39}{16}\right)=-2\left(x-\frac{5}{2}\right)^2-\frac{39}{8}\)

Vì: \(-2\left(x-\frac{5}{2}\right)^2-\frac{39}{8}\le\frac{39}{8}\forall x\)

GTLN  của bt là 39/8 tại \(-2\left(x-\frac{5}{2}\right)^2=0\Rightarrow x=\frac{5}{2}\)

cn lại lm tg tự  nha bn

\(A=15-4x^2+5x\)

\(\Rightarrow A=-4x^2+5x+15\)

\(\Rightarrow A=-4\left(x^2+\dfrac{5}{4}x+\dfrac{25}{64}\right)+\dfrac{25}{16}+15\)

\(\Rightarrow A=-4\left(x+\dfrac{5}{8}\right)^2+\dfrac{265}{16}\)

mà \(-4\left(x+\dfrac{5}{8}\right)^2\le0,\forall x\in R\)

\(\Rightarrow A=-4\left(x+\dfrac{5}{8}\right)^2+\dfrac{265}{16}\le\dfrac{265}{16}\)

\(\Rightarrow GTLN\left(A\right)=\dfrac{265}{16}\left(tại.x=-\dfrac{5}{8}\right)\)

\(A=-5x^2-4x+1\)

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

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

\(=-5\left[\left(x+\frac{2}{5}\right)^2-\frac{1}{25}\right]\)

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

Vì \(-5\left(x+\frac{2}{5}\right)^2\le0;\forall x\)

\(\Rightarrow-5\left(x+\frac{2}{5}\right)^2+\frac{1}{5}\le0+\frac{1}{5};\forall x\)

Hay \(A\le\frac{1}{5};\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{2}{5}\right)^2=0\)

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

Vậy \(A_{max}=\frac{1}{5}\Leftrightarrow x=\frac{-2}{5}\)

b: \(x^2-x+1=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

c: \(A=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi x=3

d: \(B=-\left(x^2-4x+5\right)=-\left(x^2-4x+4+1\right)=-\left(x-2\right)^2-1\le-1\forall x\)

Dấu '=' xảy ra khi x=2

a)

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

Daaus = xayr ra khi: x = 2

b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)

Dấu = xảy ra khi x = 3

c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu = xảy ra khi

2x = y và y = 2

=> x = 1 và y = 2

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

Dấu "=" <=> x = 2

b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)

Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)

c) \(4x^2+2y^2-4xy-4y+1\)

= \(\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3\)

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

Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

a) A= x² + 4x + 5

=x²+4x+4+1

=(x+2)²+1≥0+1=1

Dấu = khi x+2=0 <=>x=-2

Vậy Amin=1 khi x=-2

b) B= ( x+3 ) ( x-11 ) + 2016

=x²-8x-33+2016

=x²-8x+16+1967

=(x-4)²+1967≥0+1967=1967

Dấu = khi x-4=0 <=>x=4

Vậy Bmin=1967 <=>x=4

Bài 2:

a) D= 5 - 8x - x² 

=-(x²+8x-5)

=21-x²+8x+16

=21-x²+4x+4x+16

=21-x(x+4)+4(x+4)

=21-(x+4)(x+4)

=21-(x+4)²≤0+21=21

Dấu = khi x+4=0 <=>x=-4

Bài 1:

c)C=x²+5x+8

=x²+5x+\(\left(\dfrac{5}{2}\right)^2\)+\(\dfrac{7}{4}\)

=\(\left(x+\dfrac{5}{2}\right)^2\)+\(\dfrac{7}{4}\)\(\ge\dfrac{7}{4}\)

Vậy \(C_{min}=\dfrac{7}{4}\Leftrightarrow x=-\dfrac{5}{2}\)