Đáp án `+` Giải thích các bước giải `!`
`a)`
`3x(x-1)+x-1 = 0`
`<=> 3x(x-1)+(x-1) = 0`
`<=> (3x+1)(x-1) = 0`
`⇔` \(\left[ \begin{array}{l}3x+1=0\\x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}3x=-1\\x=1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\dfrac{1}{3}\\x=1\end{array} \right.\)
Vậy `S= {-(1)/3; 1}`
`b)`
`2(x+3)-x^2-3x = 0`
`<=> 2(x+3)-(x^2+3x) = 0`
`<=> 2(x+3)-x(x+3) = 0`
`<=> (2-x)(x+3) = 0`
`⇔` \(\left[ \begin{array}{l}2-x=0\\x+3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\)
Vậy `S= {2; -3}`
`c)`
`x^3+27+(x+3)(x-9) = 0`
`<=> (x^3+27)+(x+3)(x-9) = 0`
`<=> (x+3)(x^2-3x+9)+(x+3)(x-9) = 0`
`<=> (x+3)(x^2-3x+9+x-9) = 0`
`<=> (x+3)(x^2-2x) = 0`
`<=> x(x+3)(x-2) = 0`
`⇔` \(\left[ \begin{array}{l}x=0\\x+3=0\\x-2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=-3\\x=2\end{array} \right.\)
Vậy `S= {0; -3; 2}`
`d)`
`x^2-5x+6 = 0`
`<=> x^2-3x-2x+6 = 0`
`<=> (x^2-3x)-(2x-6) = 0`
`<=> x(x-3)-2(x-3) = 0`
`<=> (x-2)(x-3) = 0`
`⇔` \(\left[ \begin{array}{l}x-2=0\\x-3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\)
Vậy `S= {2; 3}`
