Ta có :

M = x ² - 6x = x ² - 6x + 9 - 9 = ( x - 3 ) ² - 9 \(\ge\)- 9 

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

                           \(\Leftrightarrow\)x - 3 = 0

                           \(\Leftrightarrow\)x = 3

Vậy M có giá trị nhỏ nhất = -9 khi x = 3

Ta có:

M=x²-6x=x(x-6)

A_min <=> x(x-6) đạt GTNN

mặt khác: để:M_min

thì: x>0 vì x=0=> M=0

còn x<0

=> x²-6x E N 

Vi M_min nên x bé nhất có thể

mà: 0<x=> M_min <=> x=1

Vậy M_min=1.(-5)=-5

M= x² - 6x

M= (x² - 2.3.x + 3²) - 9\(\ge\)-9

Vậy min M = -9 

1) \(M=9x^2-6x+6=\left(9x^2-6x+1\right)+5=\left(3x-1\right)^2+5\ge5\)

\(minM=5\Leftrightarrow x=\dfrac{1}{3}\)

2) \(M=5-2x-x^2=-\left(x^2+2x+1\right)+6=-\left(x+1\right)^2+6\le6\)

\(maxM=6\Leftrightarrow x=-1\)

3) \(N=5+6x-9x^2=-\left(9x^2-6x+1\right)+6=-\left(3x-1\right)^2+6\le6\)

\(maxN=6\Leftrightarrow x=\dfrac{1}{3}\)

1.(x+3)²-(x+2)(x-2)=1

x²+6x+9-(x²-4)=1

x²+6x+9-x²+4=1

6x+13=1

6x=-12

x=-2

2.Ta co: M=x²-6x

=x²-6x+9-9

=(x-3)²-9

\(\Rightarrow\)(x-3)²\(\ge\)0 với mọi x

\(\Rightarrow\)(x-3)² \(\ge\)-9

Vậy GTNN là -9

Dau "=" xảy ra khi : x-3=0=>x=3.

ĐS: GTNN = -9 <=> x=3

ĐS: GTNN = -9

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

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 "=" \(\Leftrightarrow x=2\)

\(A\left(x\right)=\dfrac{4x^4+81}{2x^2-6x+9}\)

\(=\dfrac{4x^4+36x^2+81-36x^2}{2x^2-6x+9}\)

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

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

\(=2x^2+6x+9\)

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

\(=2\left(x^2+3x+\dfrac{9}{2}\right)\)

\(=2\left(x^2+3x+\dfrac{9}{4}+\dfrac{9}{4}\right)\)

\(=2\left(x+\dfrac{3}{2}\right)^2+\dfrac{9}{2}>=\dfrac{9}{2}\forall x\)

Dấu '=' xảy ra khi \(x+\dfrac{3}{2}=0\)

=>\(x=-\dfrac{3}{2}\)

\(M=6x^2+4y^2+6xy+\left(xy+\dfrac{4x}{y}\right)+\left(3xy+\dfrac{3y}{x}\right)+2022\)

\(M\ge3x^2+y^2+3\left(x+y\right)^2+2\sqrt{\dfrac{4x^2y}{y}}+2\sqrt{\dfrac{9xy^2}{x}}+2022\)

\(M\ge3\left(x^2+1\right)+\left(y^2+4\right)+3\left(x+y\right)^2+4x+6y+2015\)

\(M\ge6x+4y+3\left(x+y\right)^2+4x+6y+2015\)

\(M\ge3\left(x+y\right)^2+10\left(x+y\right)+2015\ge3.3^2+10.3+2015=2072\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(1;2\right)\)

a) (x + 3)² - (x - 2)(x + 2) = 1

x² + 6x + 9 - x² + 4 - 1 = 0

6x + 12 = 0

6x = 0 - 12

6x = -12

x = -12/6

x = -2

b) M = x² - 6x

= x² - 6x + 9 - 9

= (x - 3)² - 9

Do (x - 3)² ≥ 0 với mọi x ∈ R

⇒ (x - 3)² - 9 ≥ -9

Vậy giá trị nhỏ nhất của M là -9 khi x = 3