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
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 （⌒▽⌒）
1) $(2x+7)(x-5)(5x+1)=0$
$⇔\left[ \begin{array}{l}2x+7=0\\x-5=0\\5x+1=0\end{array} \right.⇔\left[ \begin{array}{l}x=\frac{-7}{2}\\x=5\\x=\frac{-1}{5}\end{array} \right.$
Vậy $S=${$\frac{-7}{2};5;\frac{-1}{5}$}
2) $(x+2)(3-4x)=x^2+4x+4$
$⇔(x+2)(3-4x)=(x+2)^2$
$⇔(x+2)(1-5x)=0$
$⇔\left[ \begin{array}{l}x+2=0\\1-5x=0\end{array} \right.⇔\left[ \begin{array}{l}x=-2\\x=\frac{1}{5}\end{array} \right.$
Vậy $S=${$-2;\frac{1}{5}$}.
3) $(2x-5)^2-(x+2)^2=0$
$⇔(2x-5+x+2)(2x-5-x-2)=0$
$⇔3(x-1)(x-7)=0$
$⇔\left[ \begin{array}{l}x-1=0\\x-7=0\end{array} \right.⇔\left[ \begin{array}{l}x=1\\x=7\end{array} \right.$
Vậy $S=${$1;7$}.