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