`a)(x-2)(5-x)=0`
`→` \(\left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
Vậy `x∈{2;5}`
`b)(x^2-1)(x^2+1)=0`
Ta có: `x^2+1≥1`
`→x^2-1=0`
`→x^2=1`
`→x=±1`
Vậy `x∈{1;-1}`
`c)-12(x-5)=7(3-x)-5` `(?)`
`→-12x+60=21-7x-5`
`→-12x+7x=21-5-60`
`→-5x=-44`
`→x=44/5`
Vậy `x=44/5`
`d)30(x+2)-6(x-5)=24x`
`→30x+60-6x+30=24x`
`→30x-6x-24x=0-30-60`
`→0x=-90`
Vậy `x∈∅`