Giải thích các bước giải:
a, `x^2-4-(x-2)(x-5)=0`
`⇔(x-2)(x+2)-(x-2)(x-5)=0`
`⇔(x-2)(x+2-x+5)=0`
`⇔(x-2).7=0`
`⇔x=2`
Vậy: `S={2}`
b, `(x-2)(x+1)=x^2-4`
`⇔(x-2)(x+1)-x^2-4=0`
`⇔(x-2)(x+1)-(x-2)(x+2)=0`
`⇔(x-2)(x+1-x-2)=0`
`⇔(x-2).(-1)=0`
`⇔x=2`
Vậy: `S={2}`
c, `(x+3)(x-5)+(x+3)(3x-4)=0`
`⇔(x+3)(x-5+3x-4)=0`
`⇔(x+3)(4x-9)=0`
`⇔`\(\left[ \begin{array}{l}x+3=0\\4x-9=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-3\\x=\dfrac{9}{4}\end{array} \right.\)
Vậy: `S={-3;\frac{9}{4}}`
d, `(x+6)(3x-1)+x+6=0`
`⇔(x+6)(3x-1+1)=0`
`⇔(x+6).3x=0`
`⇔`\(\left[ \begin{array}{l}x+6=0\\3x=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-6\\x=0\end{array} \right.\)
Vậy: `S={-6;0}`
e, `(2x-7)^2-6(2x-7)(x-3)=0`
`⇔(2x-7)(2x-7-6x+18)=0`
`⇔`\(\left[ \begin{array}{l}2x-7=0\\-4x+11=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{7}{4}\\x=\dfrac{11}{4}\end{array} \right.\)
Vậy: `S={\frac{7}{4};\frac{11}{4}}`
