(x+2)^2=9*(x^2-4*x+4) 4*(2*x+7)-9*(x+3)^2=0

(x + 2)² = 9(x² – 4x + 4)

⇔ x² + 4x + 4 = 9x² – 36x + 36

⇔ x² + 4x + 4 – 9x² + 36x – 36 = 0

⇔ -8x² + 40x – 32 = 0

⇔ -8(x² – 5x + 4) = 0

⇒ x² – 5x + 4 = 0

⇔ x² – 4x – x + 4 = 0

⇔ x(x – 4) – (x – 4) = 0

⇔ (x – 4)(x – 1) = 0

⇔ \(\left[ \begin{array}{l}x-4=0\\x-1=0\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}x=4\\x=1\end{array} \right.\)

Vậy x = 4 hoặc x = 1

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4(2x + 7)² – 9(x + 3)² = 0

⇔ [2(2x + 7) – 3(x + 3)] . [2(2x + 7) + 3(x + 3)] = 0

⇔ (4x + 14 – 3x – 9) . (4x + 14 + 3x + 9) = 0

⇔ (x + 5) . (7x + 23) = 0

⇔ \(\left[ \begin{array}{l}x+5=0\\7x+23=0\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}x=-5\\x=\frac{-23}{7}\end{array} \right.\)

Vậy x = -5 hoặc x = `\frac{-23}{7}`