Đáp án:
Giải thích các bước giải:
`9x^2-1=(1-3x)(4x+1)`
`<=> (3x-1)(3x+1)= (1-3x)(4x+1)`
`<=> (3x-1)(3x+1)-(1-3x)(4x+1)=0`
`<=> (3x-1)(3x+1)+(3x-1)(4x+1)=0`
`<=> (3x-1)(3x+1+4x+1)=0`
`<=> (3x-1)(7x+2)=0`
`<=>` \(\left[ \begin{array}{l}3x-1=0\\7x+2=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=-\dfrac{2}{7}\end{array} \right.\)
`-> S={-2/7; 1/3}`
