Đáp án+giải thích các bước giải:
Bài 1:
`a,(x-1)(3x+1)=0`
`<=>` \(\left[ \begin{array}{l}x-1=0\\3x+1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=1\\3x=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=1\\x=\dfrac{-1}{3} \end{array} \right.\)
Vậy `S={1;-1/3}`
`b,(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}`
`c,(3x+1)(x^2+1)=0`
`<=>` \(\left[ \begin{array}{l}3x+1=0\\x^2+1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}3x=-1\\x^2=-1 (vô - n_0)\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{-1}{3} \\x=-1\end{array} \right.\)
Vậy phương trình có nghiệm `S={-1/3}`
`d,(3,1x-6,2)(0,5x+1)=0`
`<=>` \(\left[ \begin{array}{l}3,1x-6,2=0\\0,5x+1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}3,1x=6,2\\0,5x=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
Vậy `S={2;-2}`
