`a)(x^2-9)(3x+15)=0`
`→` \(\left[ \begin{array}{l}x^2-9=0\\3x+15=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x^2=9\\3x=-15\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=±3\\x=-5\end{array} \right.\)
Vậy `x∈{3;-3;-5}`
`b)(4x-8)(x^2+1)=0`
`→4x-8=0` (vì `x^2+1≥1`)
`→4x=8`
`→x=2`
Vậy `x=2`
`c)(x+1)(3-x)=0`
`→` \(\left[ \begin{array}{l}x+1=0\\3-x=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=-1\\x=3\end{array} \right.\)
Vậy `x∈{-1;3}`
`d)(x-2)(2x-1)=0`
`→` \(\left[ \begin{array}{l}x-2=0\\2x-1=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=2\\x=\dfrac{1}{2}\end{array} \right.\)
Vậy `x∈{2;1/2}`
`e)(3x+9)(1-3x)=0`
`→` \(\left[ \begin{array}{l}3x+9=0\\1-3x=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=-3\\x=\dfrac{1}{3}\end{array} \right.\)
Vậy `x∈{-3;1/3}`
`f)(x^2+1)(81-x^2)=0`
`→81-x^2=0` (vì `x^2+1≥1`)
`→x^2=81`
`→x^2=9^2`
`→` \(\left[ \begin{array}{l}x=9\\x=-9\end{array} \right.\)
Vậy `x∈{9;-9}`
