\(A=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow x=1\\ B=2\left(x^2-3x\right)=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ B=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ B_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\)

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

Dấu "=" \(\Leftrightarrow x=-1\)

b,\(B=2\left(x^2-3x\right)=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

Dấu "=" \(\Leftrightarrow x=\dfrac{3}{2}\)

c,\(=C=-\left(x^2-4x-3\right)=-\left[\left(x^2-4x+4\right)-7\right]=-\left(x-2\right)^2+7\le7\)

Dấu "=" \(\Leftrightarrow x=2\)

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

Đặt \(x^2+x=a\) nên \(A=a\left(a-4\right)=a^2-4a+4-4=\left(a-2\right)^2-4\ge-4\) 

Dấu "=" xảy ra \(\Leftrightarrow a-2=0\Leftrightarrow x^2+x-2=0\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

Vậy \(A_{min}=-4\) tại \(\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

B;C tương tự

ta có \(A=x^2-5x+3=x^2-\frac{2.x.5}{2}+\frac{5^2}{4}-\frac{13}{4}=\left(x-\frac{5}{2}\right)^2-\frac{13}{4}\)

vì \(\left(x-\frac{5}{2}\right)^2\ge0\Rightarrow A\ge-\frac{13}{4}\)

dáu = xảy ra <=> x=5/2

b) ta có \(B=2x^2-4x+5=2\left(x^2-2x+\frac{5}{2}\right)\) \(=2\left(x^2-2x+1+\frac{3}{2}\right)=2\left[\left(x-1\right)^2+\frac{3}{2}\right]=2\left(x-1\right)^2+3\)

vì \(2\left(x-1\right)^2\ge0\Rightarrow B\ge3\)

dấu = xảy ra <=> x=1

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

a) = 9(x² - 2.x/2.9 + 1/324) - 9/324 +5

GTNN A = 4,97

b) = (2x +y)² + y² + 2018

GTNN B = 2018 khi x=0;y=0

c) = -4(x² - 2.3x/ 4.2 + 9/16) +9/16 +10

GTLN C = 169/16

d) = -(x-y)² - (2x +1) +1 + 2016

GTLN D = 2017

(trg bn cho bài khó dữ z, làm hại cả não tui)

cảm ơn nhiều lắm đấy

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

\(A=x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)

Vậy GTNN A là 6 khi x - 2 = 0 <=> x = 2 

\(B=\left(1-x\right)\left(3x-4\right)=3x-4-3x^2+4x=-3x^2+7x-4\)

\(=-3\left(x^2-\frac{7}{3}x+\frac{4}{3}\right)=-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{1}{36}\right)=-3\left(x-\frac{7}{6}\right)^2+\frac{1}{12}\ge\frac{1}{12}\)

\(=3\left(x-\frac{7}{6}\right)^2-\frac{1}{12}\le-\frac{1}{12}\)Vậy GTLN B là -1/12 khi x = 7/6 

\(C=3x^2-9x+5=3\left(x^2-3x+\frac{5}{3}\right)=3\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{7}{12}\right)\)

\(=3\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\ge-\frac{7}{4}\)Vậy GTNN C là -7/4 khi x = 3/2 

\(D=-2x^2+5x+2=-2\left(x^2-\frac{5}{2}x-1\right)=-2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}-\frac{41}{16}\right)\)

\(=-2\left(x-\frac{5}{4}\right)^2+\frac{21}{8}\le\frac{21}{8}\)Vậy GTLN D là 21/8 khi x = 5/4 

c: \(-x^2+2x-2=-\left(x-1\right)^2-1\le-1\forall x\)

\(\Leftrightarrow V\ge-1\forall x\)

Dấu '=' xảy ra khi x=1