Đáp án + Giải thích các bước giải:
`a) 3x^2 - 4x - 12 = -8`
`⇔ 3x^2 - 4x - 4 = 0`
`⇔ (3x^2+2x)+(-6x-4) = 0`
`⇔ x(3x+2) - 2(3x+2) = 0`
`⇔ (x-2)(3x+2) = 0`
`⇔`\(\left[ \begin{array}{l}x-2=0\\3x+2=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=2\\x=-\dfrac23\end{array} \right.\)
Vậy `S = {-2/3 , 2}`
`b) 13x - 26x^2 = 9 - 18x`
`⇔ -26x^2 + 31x = 9`
`⇔ -26x^2 + 31x - 9 = 0`
`⇔ -(26x^2 - 31x + 9) = 0`
`⇔ -[(26x^2 - 13x)+(-18x+9)] = 0`
`⇔ -[13x(2x-1)-9(2x-1)] = 0`
`⇔ -(2x-1)(13x-9) = 0`
`⇔`\(\left[ \begin{array}{l}2x-1=0\\13x-9=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac12\\x=\dfrac9{13}\end{array} \right.\)
Vậy `S = {1/2,9/13}`
`c) x(x-1)(x+1)+x^2-1=0`
`⇔ x^3 + x^2 - x - 1 = 0`
`⇔ (x^3+x^2) + (-x-1) = 0`
`⇔ -(x+1) + x^2(x+1) = 0`
`⇔ (x+1)(x^2-1) = 0`
`⇔ (x+1)^2(x-1) = 0`
`⇔`\(\left[ \begin{array}{l}x+1=0\\x-1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-1\\x=1\end{array} \right.\)
Vậy `S = {-1,1}`