Đáp án `+` Giải thích các bước giải `!`
`a)`
`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}`
`b)`
`4x^2-25-(2x-5)(2x+7) = 0`
`<=> (4x^2-25)-(2x-5)(2x+7) = 0`
`<=> (2x-5)(2x+5)-(2x-5)(2x+7) = 0`
`<=> (2x-5)(2x+5-2x-7) = 0`
`<=> -2(2x-5) = 0`
`<=> 2x-5 = 0`
`<=> 2x = 5`
`<=> x = 5/2`
Vậy `S= {5/2}`
`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^3-3x^2-4x+12 = 0`
`<=> (x^3-3x^2)-(4x-12) = 0`
`<=> x^2(x-3)-4(x-3) = 0`
`<=> (x^2-4)(x-3) = 0`
`<=> (x-2)(x+2)(x-3) = 0`
`⇔` \(\left[ \begin{array}{l}x-2=0\\x+2=0\\x-3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=2\\x=-2\\x=3\end{array} \right.\)
Vậy `S= {2; -2; 3}`
