Đáp án:
`1)S={1}`
`2) S={±2}`
`3) S={0;±4}`
`4)S={0;3/2}`
Giải thích các bước giải:
`⇔1, x(1-2x)+(x-2)(2x-3)=0`
`⇔x-2x^2 + 2x^2-3x-4x+6=0`
`⇔6-6x=0`
`⇔x=1`
Vậy` S={1}`
`2, (x-1)(x+2)-x-2=0`
`(x-1)(x+2) - (x+2) =0`
`(x+2) (x-1-1)=0`
`(x+2)(x-2)=0`
`⇔`\(\left[ \begin{array}{l}x+2=0\\x-2=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-2\\x=2\end{array} \right.\)
Vậy `S={±2}`
`3, x^3-16x=0`
`⇔x(x^2-16)=0`
`⇔x(x-4)(x+4)=0`
`⇔`\(\left[ \begin{array}{l}x=0\\x-4=0\\x+4=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=±4\end{array} \right.\)
Vậy `S={0;±4}`
`4, (2x-3)^2+(x-3)(3-2x)=0`
`⇔(2x-3)^2 - (x-3)(2x-3)=0`
`⇔(2x-3)(2x-3-x+3)=0`
`⇔x(2x-3)=0`
`⇔`\(\left[ \begin{array}{l}x=0\\2x-3=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=\dfrac{3}{2} \end{array} \right.\)
Vậy `S={0;3/2}`
