E = 3( x² - 5x/4 + 25/4) -25/4 +1

GTNN E = -21/4

Ta có: \(3x^2+5x+4=3\left(x+\frac{5}{6}\right)^2+\frac{23}{12}\ge\frac{23}{12}\)

Vậy..

a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42

\(x^2+\left(s-3x\right)^2-5x-15\left(s-3x\right)+8\le0\)

\(S=3x+y\Leftrightarrow y=S-3x\)

\(10x^2-2\left(3x-20\right)x+s^2-15s+8\le0\left(1\right)\)

Tìm đk S để có BPT (1) có nghiệm

Ta có:

\(\left(3s-20\right)^2-10s^2+150s-80\ge0\)

\(s^2-30s-320\le0\)

\(15-\sqrt{545}\le s\le15+\sqrt{545}\)

Vậy MinS = \(15-\sqrt{545}\)

`A=x^2-2x+5`

`=x^2-2x+1+4`

`=(x-1)^2+4>=4`

Dấu "=" `<=>x=1`

`B=4x^2+4x+3`

`=4x^2+4x+1+2`

`=(2x+1)^2+2>=2`

Dấu "=" xảy ra khi `x=-1/2`

`C=9x^2-6x+7`

`=9x^2-6x+1+6`

`=(3x-1)^2+6>=6`

Dấu '=' xảy ra khi `x=1/3`

`D=5x^2+3x+8`

`=5(x^2+3/5x)+8`

`=5(x^2+3/5x+9/100-9/100)+8`

`=5(x+3/10)^2+151/20>=151/20`

Dấu "=" xảy ra khi `x=-3/10`

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

Ta có: \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\Rightarrow A_{min}=4\) khi \(x=1\)

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

Ta có: \(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+2\ge2\Rightarrow B_{min}=2\) khi \(x=-\dfrac{1}{2}\)

\(C=9x^2-6x+7=9x^2-6x+1+6=\left(3x-1\right)^2+6\)

Ta có: \(\left(3x-1\right)^2\ge0\Rightarrow\left(3x-1\right)^2+6\ge6\Rightarrow C_{min}=6\) khi \(x=\dfrac{1}{3}\)

\(D=5x^2+3x+8\Rightarrow5\left(x^2+2.x.\dfrac{3}{10}+\dfrac{9}{100}\right)+\dfrac{151}{20}=5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\)

Ta có: \(5\left(x+\dfrac{3}{10}\right)^2\ge0\Rightarrow5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\ge\dfrac{151}{20}\)

\(\Rightarrow D_{min}=\dfrac{151}{20}\) khi \(x=-\dfrac{3}{10}\)

- A = (x-1)² + 4 \(\ge4\)

Dấu "=" <=> x = 1

- B = (2x+1)² +2 \(\ge2\)

Dấu "=" xảy ra <=> x = \(\dfrac{-1}{2}\)

- C = (3x - 1)² + 6 \(\ge6\)

Dấu "=" <=> x = \(\dfrac{1}{3}\)

- D = \(5\left(x^2+\dfrac{3}{5}x+\dfrac{9}{100}\right)+\dfrac{151}{20}=5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\ge\dfrac{151}{20}\)

Dấu "=" <=> x = \(\dfrac{-3}{10}\)

\(A=x^2+9x+56=\left(x+\frac{9}{2}\right)^2+\frac{143}{4}\)

Vì \(\left(x+\frac{9}{2}\right)^2\ge0\forall x\)\(\Rightarrow\left(x+\frac{9}{2}\right)^2+\frac{143}{4}\ge\frac{143}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{9}{2}\right)^2=0\Leftrightarrow x=-\frac{9}{2}\)

Vậy minA = 143/4 <=> x = - 9/2

\(B=x^2-2x+15=\left(x-1\right)^2+14\)

Vì \(\left(x-1\right)^2\ge0\)\(\Rightarrow\left(x-1\right)^2+14\ge14\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)

Vậy minB = 14 <=> x = 1

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

Vì \(\left(x-\frac{2}{3}\right)^2\ge0\forall x\)\(\Rightarrow9\left(x-\frac{2}{3}\right)^2-4\ge-4\)

Dấu "=" xảy ra \(\Leftrightarrow9\left(x-\frac{2}{3}\right)^2=0\Leftrightarrow x-\frac{2}{3}=0\Leftrightarrow x=\frac{2}{3}\)

Vậy minC = - 4 <=> x = 2/3

Bài 1.

A = x² + 9x + 56

= ( x² + 9x + 81/4 ) + 143/4

= ( x + 9/2 )² + 143/4

( x + 9/2 )² ≥ 0 ∀ x => ( x + 9/2 )² + 143/4 ≥ 143/4

Đẳng thức xảy ra <=> x + 9/2 = 0 => x = -9/2

=> MinA = 143/4 <=> x = -9/2

B = x² - 2x + 15

= ( x² - 2x + 1 ) + 14

= ( x - 1 )² + 14

( x - 1 )² ≥ 0 ∀ x => ( x - 1 )² + 14 ≥ 14 

Đẳng thức xảy ra <=> x - 1 = 0 => x = 1

=> MinB = 14 <=> x = 1 

C = 9x² - 12x 

= 9( x² - 4/3x + 4/9 ) - 4

= 9( x - 2/3 )² - 4

9( x - 2/3 )² ≥ 0 ∀ x => 9( x - 2/3 )² - 4 ≥ -4

Đẳng thức xảy ra <=> x - 2/3 = 0 => x = 2/3

=> MinC = -4 <=> x = 2/3

Bài 2.

D = -9x² + x

= -9( x² - 1/9x + 1/324 ) + 1/36

= -9( x - 1/18 )² + 1/36

-9( x - 1/18 )² ≤ 0 ∀ x => -9( x - 1/18 )² + 1/36 ≤ 1/36

Đẳng thức xảy ra <=> x - 1/18 = 0 => x = 1/18

=> MaxD = 1/36 <=> x = 1/18

E = -x² + 3x - 5

= -( x² - 3x + 9/4 ) - 11/4

= -( x - 3/2 )² - 11/4

-( x - 3/2 )² ≤ 0 ∀ x => -( x - 3/2 )² - 11/4 ≤ -11/4

Đẳng thức xảy ra <=> x - 3/2 = 0 => x = 3/2

=> MaxE = -11/4 <=> x = 3/2

F = -16x² - 5x

= -16( x² + 5/16x + 25/1024 ) + 25/64

= -16( x + 5/32 )² + 25/64 

-16( x + 5/32 )² ≤ 0 ∀ x => -16( x + 5/32 )² + 25/64 ≤ 25/64

Đẳng thức xảy ra <=> x + 5/32 = 0 => x = -5/32

=> MaxF = 25/64 <=> x = -5/32