Đáp án+Giải thích các bước giải:
Bài 5:
1)
x(x-5)-2(x-5)=0
⇔(x-5)(x-2)=0
⇔\(\left[ \begin{array}{l}x-5=0\\x-2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=5\\x=2\end{array} \right.\)
Vậy S={5,2}
2)
3x(x-4)-x+4=0
⇔3x(x-4)-(x-4)=0
⇔(x-4)(3x-1)=0
⇔\(\left[ \begin{array}{l}x-4=0\\3x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=4\\x=\frac{1}{3}\end{array} \right.\)
Vậy S={4,`(1)/(3)`}
3)
x(x-7)-2(7-x)=0
⇔x(x-7)+2(x-7)=0
⇔(x-7)(x+2)=0
⇔\(\left[ \begin{array}{l}x-7=0\\x+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=7\\x=-2\end{array} \right.\)
Vậy S={7,-2}
4)
2x(2x+3)-2x-3=0
⇔2x(2x+3)-(2x+3)=0
⇔(2x+3)(2x-1)=0
⇔\(\left[ \begin{array}{l}2x+3=0\\2x-1=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=\frac{-3}{2}\\x=\frac{1}{2}\end{array} \right.\)
Vậy S={`(-3)/(2)`,`(1)/(2)`}
