Giải các phương trình sau (phương trình tích) 1) (2x + 7)(x – 5)(5x + 1) = 0 2) (x + 2)(3 – 4x) = x² + 4x + 4 3) (2x – 5)² – (x + 2)² = 0

$1) (2x + 7)(x – 5)(5x + 1) = 0$

$⇔$ \(\left[ \begin{array}{l}2x + 7=0\\x-5=0\\5x+1=0\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}2x=-7 \to x=\dfrac{-7}{2}\\x –5=0 \to x=5\\5x=-1 \to x=\dfrac{-1}{5}\end{array} \right.\) 

$2) (x + 2)(3 – 4x) = x² + 4x + 4$

$⇔(x+2)(3-4x)=(x+2)²$

$⇔(x+2)(3-4x)-(x+2)²=0$

$⇔(x+2)(3-4x-x-2)=0$

$⇔(x+2)(1-5x)=0$

$⇔$\(\left[ \begin{array}{l}x+2=0\\1-5x=0\end{array} \right.\)

$⇔$ \(\left[ \begin{array}{l}x=-2\\x=\dfrac{1}{5}\end{array} \right.\)

$3) (2x – 5)² – (x + 2)² = 0$

$⇔ ( 2 x − 5 + x + 2 ) ( 2 x − 5 − x − 2 ) = 0$

$⇔ ( 3x − 3 ) ( x − 7 ) = 0$

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

$⇔$ \(\left[ \begin{array}{l}x=1\\x=7\end{array} \right.\) 

Cho Mình Câu Trả Lời Hay Nhất Nhé （⌒▽⌒）