`\text{3)}`
`\text{1.}`
`3x^3 -12x =0`
`=> 3x(x^2 - 4) =0`
`=>` \(\left[ \begin{array}{l}3x =0\\x^2-4=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=0\\x=±2 \end{array} \right.\)
Vậy `x \in {0 ;±2}`
$\\$
`2x(3x-5)= 10 - 6x`
`=> 6x^2 - 10x =10 -6x`
`=> 6x^2 -10x - 10 +6x = 0`
`=> 6x^2 - 4x - 10 =0`
`=> 6(x^2 - 2 . 1/3x - 5/3) = 0`
`=> 6(x^2 - 2 . x . 1/3 + 1/9 -16/9) =0`
`=> (x-1/3)^2 = 16/9`
`=> (x-1/3)^2 = 4/3`
`=>` \(\left[ \begin{array}{l}x-\dfrac{1}{3} = \dfrac{4}{3}\\x-\dfrac{1}{3} = \dfrac{-4}{3}\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=\dfrac{5}{3}\\x=-1\end{array} \right.\)
Vậy `x \in {5/3 ;-1}`
$\\$
`\text{2.}`
`(5n-2)^2 - (2n-5)^2`
`= (5n-2 + 2n -5)(5n-2 - 2n +5 )`
`= (7n- 7)(3n +3 )`
`= 7(n-1) . 3(n+1)`
`= 21(n-1)(n+1) \vdots 21`
Vậy `(5n-2)^2 - (2n-5)^2` chia hết cho `21` với mọi `n \in Z`
