~ Bạn tham khảo ~
`1, x^3 - 16x = 0`
`<=> x(x^2 - 16) = 0`
`<=> x(x-4)(x+4)=0`
\(⇔\left[ \begin{array}{l}x=0\\x-4=0\\x+4=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=0\\x=4\\x=-4\end{array} \right.\)
Vậy `S={0;4;-4}`
`2, x^2-25x=0`
`<=> x(x-25)=0`
\(⇔\left[ \begin{array}{l}x=0\\x-25=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=0\\x=25\end{array} \right.\)
Vậy `S={0;25}`
`3,(3x-5)^2 -( x-4)^2 =0`
`<=> (3x-5-x+4)(3x-5+x-4)=0`
`<=> (2x-1)(4x-9)=0`
\(⇔\left[ \begin{array}{l}2x-1=0\\4x-9=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}2x=1\\4x=9\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=\dfrac12\\x=\dfrac94\end{array} \right.\)
Vậy `S={1/2;9/4}`
`4, 25x^2 =(4x-3)^2`
`<=> (5x)^2 - (4x-3)^2=0`
`<=> (5x-4x+3)(5x+4x-3)=0`
`<=> (x+3)(9x-3)=0`
`<=> 3(x+3)(3x-1)=0`
\(⇔\left[ \begin{array}{l}x+3=0\\3x-1=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=-3\\3x=1\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=-3\\x=\dfrac13\end{array} \right.\)
Vậy `S={-3;1/3}`
