Đáp án:
`b)` `S={\frac{-2}{3}; 1}`
`d)` `S={2; \frac{-1}{2}}`
`f)` `S={2;-3}`
Giải thích các bước giải:
`b)` `3x(x-1)+2(x-1)=0`
`<=> (x-1)(3x+2)=0`
`<=> \(\left[ \begin{array}{l}x-1=0\\3x+2=0\end{array} \right.\) `<=> ` \(\left[ \begin{array}{l}x=1\\x=-\frac{2}{3}\end{array} \right.\)
Vậy `S={-\frac{2}{3}; 1}`
`d)` `2x(x-2)+x-2=0`
`<=> 2x(x-2)+(x-2)=0`
`<=> (x-2)(2x+1)=0`
`<=> `\(\left[ \begin{array}{l}x-2=0\\2x+1=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=2\\x=\frac{-1}{2}\end{array} \right.\)
Vậy `S={-\frac{1}{2}; 2}`
`f)` `2(x+3)-x^2-3x=0`
`<=> 2(x+3)-x(x+3)=0`
`<=> (x+3)(2-x)=0`
`<=>` \(\left[ \begin{array}{l}x+3=0\\2-x=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-3\\x=2\end{array} \right.\)
Vậy `S={2; -3}`
