Đáp án:

$1,$

$a, x² + 6xy + 9y² $

$= x² + 2.3xy + (3y)² $

$= (x + 3y)² $

$b, 25 - 20x + 16x² $

$= 5² - 2.5.4x + (4x)² $

$= (5 - 4x)² $

$c, (2x + 3y)(2x - 3y) $

$= (2x)² - (3y)² $

$= 4x² - 9y² $

$d, x² + 2x + 1 $

$= (x + 1)² $

$e, 9x² + y² + 6xy $

$= 9x² + 6xy + y² $

$= (3x + y)² $

$f, x² - x + \frac{1}{4} $

$= x² - 2.\frac{1}{2}x + \frac{1}{4} $

$= (x - \frac{1}{2})² $

$2,$

$a,  (x - 2)² - 25 = 0 $

$⇔ (x - 2)² - 5² = 0 $

$⇔ (x - 2 - 5)(x - 2 + 5) = 0 $

$⇔ (x - 7)(x + 3) = 0 $

$⇔$ \(\left[ \begin{array}{l}x-7=0\\x+3=0\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}x=7\\x=-3\end{array} \right.\) 

$b, (x² + 6x + 9) - (4x² + 4x + 1) = 0 $

$⇔ (x + 3)² - (2x + 1)² = 0 $

$⇔ (x + 3 - 2x - 1)(x + 3 + 2x + 1) = 0 $

$⇔ (2 - x)(3x + 4) = 0 $

$⇔$ \(\left[ \begin{array}{l}2-x=0\\3x+4=0\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}x=2\\x=\frac{-4}{3}\end{array} \right.\) 

$c, x² + 4x + 3 = 0 $

$⇔ x² + 4x + 4 - 1 = 0 $

$⇔ (x + 2)² - 1 = 0 $

$⇔ (x + 2 - 1)(x + 2 + 1) = 0 $

$⇔ (x + 1)(x + 3) = 0 $

$⇔$ \(\left[ \begin{array}{l}x+1=0\\x+3=0\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}x=-1\\x=-3\end{array} \right.\) 

$d, x² - 6x + 8  = 0 $

$⇔ x² - 6x + 9 - 1 = 0 $

$⇔ (x - 3)² - 1 = 0 $

$⇔ (x - 3 - 1)(x - 3 + 1) = 0 $

$⇔ (x - 4)(x - 2) = 0 $

$⇔$ \(\left[ \begin{array}{l}x-4=0\\x-2=0\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}x=4\\x=2\end{array} \right.\) 

$3,$

$A = 4x² + 4x + 15 $

$A = 4x² + 4x + 1 + 14 $

$A = (2x + 1)² + 14 ≥ 14  $

$Amin = 14 ⇔ (2x + 1)² = 0 $

                 $ ⇔ x = \frac{-1}{2} $

$B = 9x² + 6x - 5 $

$B = 9x² + 6x + 1 -6 $

$B = (3x + 1)² -6  ≥ -6 $

$Bmin = -6 ⇔ (3x - 1)² = 0$

               $⇔ x = \frac{-1}{3}$