ta có 

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

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

( 2x + 1 )^2 - ( 2x + 1 ) ( 2x - 4 ) + ( x - 2 )^2

=  ( 2x + 1 )^2 - 2 ( 2x + 1 ) ( x - 2 ) + ( x - 2 )^2

= ( 2x + 1 - x + 2 )^2

= ( x + 3 )^2

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

\(=4x^2+4x+1-\left(4x^2-6x-4\right)+x^2-4x+4\)

\(=5x^2+5-4x^2+6x+4=x^2+6x+9=\left(x+3\right)^2\)

a * 1 ( 1 + 3) - 3 ( 1 -3)

=  a * 4 + 6 

= 4a + 6

`ax (x-y) + y (x+y)`

`= ax^2 - axy  + xy + y^2`

`= a . 1^2 - a . 1 . (-3) + 1 . (-3)+(-3)^2`

`= a  + 3a - 3 + 9`

`= 4a - 6`

`3x^2y (5x + 2y-4)`

`= 15x^3y + 6x^2y^2 - 12x^2y`

a)\(x^2-2.x.3+3^2+2x-x^2=-7\)

\(-4x+9=-7\)

\(-4x=-16=>x=4\)

b)\(12x-4x^2+\left[\left(2x\right)^2-1^2\right]=5\)

\(12x-4x^2+4x^2-1=5\)

\(12x-1=5\)

\(12x=6=>x=\frac{1}{2}\)

\(a,\left(x-3\right)^2+\left(x+3\right)^2-\left(x-2\right)\left(x+2\right)\)

\(=x^2-6x+9+x^2+6x+9-x^2+4=x^2+22\)

\(b,x^2+22=|3|^2+22=9+22=31\)

\(c,x^2+22\)

Vì \(x^2\ge0\)   \(\Rightarrow\)  \(x^2+22\ge22\left(\forall x\right)\)     

Dấu ''='' xảy ra   \(\Leftrightarrow\)  \(x^2=0\)     

                           \(\Leftrightarrow\)    \(x=0\)

 vậy giá trị nhỏ nhất của  A  là:

                         Min A  =  22   khi  x  = 0