Đáp án + Giải thích các bước giải:
`a)`
`x^{2}-4x=0`
`<=>x(x-4)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\x-4=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=0\\x=4\end{array} \right.\)
Vậy `S={0;4}`
`b)`
`x^{2}+10x=-25`
`<=>x^{2}+10x+25=0`
`<=>(x+5)^{2}=0`
`<=>x=-5`
Vậy `S={-5}`
`c)`
`8x^{3}+12x^{2}+6x+1=0`
`<=>(2x)^{3}+3.(2x)^{2}.1+3.2x.1^{2}+1^{3}=0`
`<=>(2x+1)^{3}=0`
`<=>x=-1/2`
Vậy `S={-1/2}`
`d)`
`(x-5)^{2}-1=0`
`<=>(x-5-1)(x-5+1)=0`
`<=>(x-6)(x-4)=0`
`<=>` \(\left[ \begin{array}{l}x-6=0\\x-4=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=6\\x=4\end{array} \right.\)
Vậy `S={6;4}`
`e)`
`(x+3)^{2}-(x-2)^{2}=0`
`<=>(x+3-x+2)(x+3+x-2)=0`
`<=>5(2x+1)=0`
`<=>2x+1=0`
`<=>x=-1/2`
Vậy `S={-1/2}`
