Đáp án:
Giải thích các bước giải:
`4)`
`a)(x-1)(x+2)-x-2=0`
`<=>(x-1)(x+2)-(x+2)=0`
`<=>(x-1-1)(x+2)=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 phương trình có tập nghiệm `{2;-2}`
`b)x(2x-3)-2(3-2x)=0`
`<=>x(2x-3)+2(2x-3)=0`
`<=>(x+2)(2x-3)=0`
`<=>`\(\left[ \begin{array}{l}x+2=0\\2x-3=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-2\\x=3/2\end{array} \right.\)
`c)(3x-2)^2=(4x+1)^2`
`<=>9x^2-12x+4-16x^2-8x-1=0`
`<=>-7x^2-20x+3=0`
`<=>-7x^2-21x+x+3=0`
`<=>-7x(x+3)+(x+3)=0`
`<=>(-7x+1)(x+3)=0`
`<=>`\(\left[ \begin{array}{l}-7x+1=0\\x+3=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=1/7\\x=-3\end{array} \right.\)
`d)x^2-5x+6=0`
`<=>x^2-2x-3x+6=0`
`<=>(x^2-2x)-(3x-6)=0`
`<=>x(x-2)-3(x-2)=0`
`<=>(x-3)(x-2)=0`
`<=>`\(\left[ \begin{array}{l}x-3=0\\x-2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=3\\x=2\end{array} \right.\)
`e)3x^2+x-2=0`
`<=>3x^2+3x-2x-2=0`
`<=>3x(x+1)-2(x+1)=0`
`<=>(3x-2)(x+1)=0`
`<=>`\(\left[ \begin{array}{l}3x-2=0\\x+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=2/3\\x=-1\end{array} \right.\)
`f)x^3+27+(x+3)(x-9)=0`
`<=>x^3+3^3+(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+3)x(x-2)=0`
`<=>`\(\left[ \begin{array}{l}x+3=0\\x(x-2)=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-3\\x=0;x=2\end{array} \right.\)
Vậy phương trình có tập nghiệm `S={0;-3;2}`
